Notes on Linear Algebra Mark Reeder December 19, 2015 Contents I The Tensor Algebra and its quotients 3 1 The Tensor Algebra 3 1.1 Filtered and graded algebras ............................... 4 2 The Symmetric Algebra 5 2.1 Polynomials and differential operators .......................... 5 3 The Exterior Algebra 6 3.1 Free modules ....................................... 7 3.2 Determinants ....................................... 9 3.3 The trace ......................................... 9 3.4 Interpolating between trace and determinant ....................... 10 3.5 Vector spaces ....................................... 11 3.5.1 The characteristic polynomial .......................... 11 3.5.2 A topological interpretation ........................... 13 3.5.3 Grassmannians .................................. 14 3.5.4 Oriented Grassmannians and the characteristic polynomial ........... 15 1
I assume the reader has learned about tensor products, as found (for example) in Dummit and Foote [DFhereafter]. Throughout, R is a commutative ring with identity and M is an R-module. We will discussfour important R-algebras constructed from M and perhaps some additional structure on M . All ofthese algebras will be quotients of a universal algebra, the tensor algebra of M . All tensor productsare taken in the category of R-modules and we write ⊗ for ⊗R. After each of the four constructions Iwill try to show why they are important and interesting.
1 The Tensor Algebra
Let T kM denote the k-fold tensor product of M with itself. Thus,
One checks that this is well-defined. Thus TM becomes an R-algebra. It is a graded algebra, in thesense that
T kM · T `M ⊂ T k+`M.
We have a natural inclusion ι : M = T 1M ↪→ TM .
The mapping property for TM is as follows. Suppose A is an R-algebra. Then any R-module mapφ : M → A extends to a unique R-module map φ : TM → A, such that φ ◦ ι = φ. This follows fromthe mapping property of tensor products.
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If M is a free R-module with basis {xi : i ∈ I}, then TM is also a free R-module, with basis{xi1 ⊗ xik : allij ∈ I}. We can think of elements of TM as noncommuting polynomial expressionsin the xi’s.
1.1 Filtered and graded algebras
An R-algebra A is graded by a semigroup K if
A =⊕k∈K
Ak
where the Ak are R-submodules of A such that Aj · Ak ⊂ Aj+k for all j, k ∈ K.
For example the tensor algebra T (V ) is evidently graded by N = {0, 1, 2, . . . }. Gradings by Z/mZ(for various m) also arise in nature.
An N-graded algebraA is filtered if there areR-submodulesAn ⊂ A, for n ∈ N, such thatA0 ⊂ A1 ⊂A2 ⊂ · · · with
A =⋃n∈N
An,
such that Am · An ⊂ Am+n for all m,n ∈ N.
Every N- graded algebra A = ⊕k∈NAk is filtered via
An =n⊕k=0
Ak. (1)
There are filtered algebras that do not arise in this way. However, given a filtered algebra A = ∪An,we can define the R-module
gr(A) =⊕k∈N
gr(A)k,
where gr(A)0 = A0 and gr(A)k = Ak/Ak−1 for k > 0. One checks that the product in A induces awell-defined product in gr(A), making gr(A) into an N-graded algebra, called the associated gradedalgebra of the filtered algebra A.
In what follows we will consider quotients of the tensor algebra T (V ) by certain two-sided ideals.Though T (V ) is graded, these quotients will not always be graded, but they will always be filtered.
Let A = ⊕Ak be an N-graded algebra, let I ⊂ A be a (two-sided) ideal in A, and let π : A→ A/I bethe projection. The R-algebra A/I is filtered by the image of the filtration of A:
(A/I)n = π(An),
with An as in (1).
We say that I is homogeneous if I is generated by elements lying in the graded subspaces Ak. Thismeans I = ⊕k(I ∩ Ak), so A/I inherits the grading
(A/I)k = Ak/I ∩ Ak, k ∈ N. (2)
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In the first of our two quotients ofA = T (V ) below, the ideal I will be homogeneous, so these quotientswill be graded.
However in the third and fourth quotients, I will not be homogeneous: there will be elements a ∈ I ofthe form a = ak + ak−1 + · · · + a0 with all aj ∈ Aj , but not all aj ∈ I . Assume ak 6= 0. The leadingterm σ(a) = ak is called the symbol of a. 1 Let σ(I) be the ideal generated by {σ(a) : a ∈ I}. Thenσ(I) is homogenous, so the quotient A/σ(I) is naturally graded, as in (2), and we have
A/σ(I) ' gr(A/I), (3)
as graded algebras, via the map sending the class of a ∈ Aj to the class of a+ I ∈ (A/I)j .
2 The Symmetric Algebra
This is the first of our four quotients of T (M). Let Isym(M) be the (two-sided) ideal in T (M) generatedby {x⊗ y − y ⊗ x : x, y ∈M}. The Symmetric Algebra of M is defined as
Sym(M) = T (M)/Isym(M).
The relations in Isym(M) force Sym(M) to be commutative. Since Isym(M) is homogeneous, thequotient Sym(M) is graded:
Sym(M) =⊕k≥0
Symk(M),
where Symk(M) is the image of T k(M) in Sym(M). We write x1x2 · · ·xk ∈ Symk(M) for the imageof x1 ⊗ · · · ⊗ xk ∈ T (M).
The mapping property for Sym(M) is as follows. Suppose A is a commutative R-algebra. Then anyR-module map φ : M → A extends to a unique R-module map φ : Sym(M)→ A, such that φ◦ ι = φ.This follows from the mapping property of T (M). For x1, . . . , xk ∈M we have
φ(x1x2 · · ·xk) = φ(x1)φ(x2) · · ·φ(xk).
In this sense, Sym(M) is the largest commutative quotient of T (M).
2.1 Polynomials and differential operators
Suppose M is free over R of rank n and that {x1, . . . , xn} is an R-basis of M . Let R[X1, . . . , Xn] bethe polynomial ring in indeterminants Xi. Since M is free, there is a unique R-module mapping
φ : M → R[X1, . . . , Xn], xi 7→ Xi.
1One can also define the symbol assuming only that A is filtered; here σ(a) is the class of a in Ak/Ak−1. However, wewill only need the symbol for ideals in graded algebras A.
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By the mapping property this extends to an isomorphism of graded algebras:
φ : Sym(M)→ R[X1, . . . , Xn].
This is surjective because the Xi generate R[X1, . . . , Xn] and is injective because the monomialsX i1
1 · · ·X inn form an R-basis of R[X1, . . . , Xn]. It follows that Sym(M) is free over R with basis
{xi11 · · ·xinn : ij ≥ 0}. Thus, Sym(M) is a coordinate-free version of the polynomial ring.
To express Sym in terms of polynomial functions, it is more natural to pass to the dual space M =HomR(M,R) of R-linear functionals on M . Let F(M) be the R-algebra of all R-valued functions onM . The inclusion ι : M ↪→ F(M) extends, by the mapping property, to an R-algebra homomorphismι : S(M) → F(M). A polynomial function on M is a element f ∈ F(M) of the form f = ι(s) forsome s ∈ S(M). Note that the mapping ι need not be injective. For example, if R = Fp = M , and0 6= x ∈ M , then ι(xp − x) = 0. Thus, a nonzero polynomial can give the zero function.
However, if R is an infinite field, then ι is injective, so we can identify Sym(M) with polynomialfunctions onM . In this interpretation, Sym(M) becomes a ring of differential operators on Sym(M),as follows. Each x ∈ M may be viewed as a linear function x : M → R λ 7→ 〈λ, x〉. This extends tothe linear map
∂x ∈ EndR(T (M))
given by
∂x(λ1 ⊗ · · · ⊗ λk) =k∑i=1
〈λi, x〉λ1 ⊗ · · ·λi−1 ⊗ λi+1 ⊗ λk.
Thus, ∂x is the unique extension satisfying the product rule. Such an endomorphism of an algebra iscalled a derivation, because ∂x is analogue of the directional derivative in Calculus.
One checks that for x, y ∈ M we have ∂x∂y = ∂y∂x. By the mapping property, x 7→ ∂x extends to analgebra homomorphism
Thus Sym(M) acts on Sym(M) via differential operators. Of course Sym(M) acts on itself by multi-plication. These two actions do not commute with each other; one checks that
∂xλ− λ∂x = 〈λ, x〉 ∈ End(Sym(M)).
This is an algebraic version of the Uncertainty Principle of Physics, in which λ detects position and ∂xdetects momentum. The subalgebra of End(Sym(M)) generated by Sym(M) and Sym(M) is calledthe Weyl Algebra. The Weyl algebra can also be described directly as a quotient of a tensor algebra,but we shall not do this.
3 The Exterior Algebra
The exterior algebra Λ(M) is the R-algebra obtained as the quotient of T (M) by the homogeneousideal Ialt(M) generated by {m ⊗m : m ∈ M}. Each graded component Λk(M) is generated as an
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R-module by the images m1 ∧ · · · ∧ mk of the k-fold pure tensors m1 ⊗ · · · ⊗ mk in T k(M). Moreprecisely we have [DF p.447]
Λk(M) = T k(M)/Ak(M),
whereAk(M) is the submodule of T k(M) generated by all k-fold tensorsm1⊗· · ·⊗mk wheremi = mj
for some 1 ≤ i < j ≤ k. Hence we have the following mapping property:
If N is an R-module and f : T k(M) → N is an R-module map vanishing on pure k-tensors with twoequal components then there is a unique R-module map
f : ΛkM → N such that f(m1 ∧ · · · ∧mk) = f(m1 ⊗ · · · ⊗mk).
By definition, we have m ∧m = 0 for all m ∈M . Expanding 0 = (m+m′) ∧ (m+m′) shows that
m ∧m′ = −m′ ∧m
for all m,m′ ∈M . It follows that switching any pair mi,mj in a pure wedge changes the sign:
If−1 = 1 in R then A(M) properly contains the ideal in T (M) generated by all expressions m⊗m′+m′ ⊗m so these do not suffice to define the relations in Λ(M).
3.1 Free modules
If M and M are free R-modules with bases {mi} and {m′j} respectively, then M ⊗RM ′ is again free,with basis {mi ⊗m′j}.
Assume M is free of rank n, and let {m1, . . . ,mn} be a basis of M . By induction, it follows thatT k(M) is free with basis {mi1 ⊗ · · · ⊗ mik} indexed by [1, n]k. Since mi1 ⊗ · · · ⊗ mik ∈ Ak(M)whenever ip = iq for p 6= q, it follows that Λk(M) is spanned by elements of the form mi1 ∧ · · · ∧mik
for 1 ≤ i1 < i2 < · · · < ik ≤ n. In particular Λk(M) = 0 for k > n and Λn(M) is a cyclic R-modulegenerated by the n-fold wedge product m1 ∧ · · · ∧mn.
Lemma 3.1 Let M be a free R-module of rank n. Then Λn(M) is a free R-module of rank one,generated by m1 ∧ · · · ∧mn, where {mi} is any basis of M .
Proof: Our remarks above show that m1 ∧ · · · ∧ mn spans Λn(M). We will prove the lemma byconstructing an R-module map ϕ : Λn(M)→ R such that ϕ(m1 ∧ · · · ∧mn) = 1. 2
Let {λ1, . . . , λn} ⊂ HomR(M,R) be the dual basis of {mi}, defined by λi(mi) = 1 and λi(mj) = 0 ifi 6= j. Then we have a well-defined R-module map Let φ : T n(M)→ R given by
φ(x1 ⊗ · · · ⊗ xn) =∑σ∈Sn
sgn(σ)λ1(xσ1) · · ·λn(xσn).
2This idea comes from P. Garrett’s notes http://www.math.umn.edu/ garrett/m/algebra/notes/28.pdf
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We show that φ vanishes on An(M): Suppose xi = xj for some i 6= j, and let τ ∈ Sn be thetransposition switching i and j. Then the sum over Sn is a sum over pairs σ, τσ:
Define p, q ∈ [1, n] by σp = i and σq = j. Then (4) becomes
sgn(σ)∏6=p,q
λ`(xσ`) · [λp(xσp)λq(xσq)− λp(xτσp)λq(xτσq)]
and the term in [· · · ] isλp(xi)λq(xj)− λp(xj)λq(xi) = 0,
since xi = xj . Hence φ induces a well-defined map ϕ : Λn(M)→ R, given by
ϕ(x1 ∧ · · · ∧ xn) =∑σ∈Sn
sgn(σ)λ1(xσ1) · · ·λn(xσn).
Taking each xi = mi, every term in the sum vanishes except for σ = e, so
ϕ(m1 ∧ · · · ∧mn) = 1
as claimed. �
We set [1, n] = {1, 2, . . . , n}, let k ∈ [1, n] and let [1, n]k be the set of k-element subsets of [1, n].Given a set of vectors {v1, . . . , vn} in V and a subset I ⊂ [1, n]k, let vI = vi1 ∧ · · · ∧ vik , wherei1 < i2 < · · · < ik are the elements of I listed in increasing order.
Proposition 3.2 If {m1, . . . ,mn} is a basis of M then {mI : I ∈ [1, n]k} is a basis of Λk(M). Inparticular Λk(M) is free of rank
(nk
).
Proof: 3
Since the k-fold pure tensor products of the mi span T k(M), the corresponding wedge products spanΛk(M). Since these wedge products may be permuted up to sign, it follows that {MI : I ∈ [1, n]k}spans Λk(M).
Suppose we have a dependence relation ∑I∈[1,n]k
cIMI = 0.
Choose J ∈ [1, n]k arbitrarily and let J ′ ∈ [1, n]n−k be its complement. Then mJ ′ ∧ mJ = ±m1 ∧· · · ∧mn, and for I 6= J ′ there exists i ∈ I ∩ J so we have mJ ∧mI = 0. Therefore
0 = mJ ′ ∧∑
I∈[1,n]k
cImI = ±cJm1 ∧ · · · ∧mn.
Since m1 ∧ · · · ∧mn 6= 0 by Lemma 3.1, it follows that cJ = 0. This shows that {mI : I ∈ [1, n]k} islinearly independent.
�3The proof of this result in [DF] is not valid if 2 = 0 in R. The proof here works in any characteristic.
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3.2 Determinants
The exterior product Λk(·) is a functor. That is, given two R-modules M,N and an R-module mapT : M → N we get for each k ≥ 1 a well-defined R-module map
In particular each T ∈ EndR(M) gives a series of R-linear maps Λk(T ) ∈ EndR(Λk(M)), for k ≥ 1.
If M is free of rank n over R, then according to Lemma 3.1, Λn(M) is a cyclic R-module generatedby the n-fold wedge product
m1 ∧ · · · ∧mn
of any basis {m1, . . . ,mn}. Therefore Λn(T ) acts on Λn(M) via multiplication by some element of R,called the determinant of T , denoted by det(T ). That is,
Λn(T ) = det(T ) · id on Λn(M).
This definition makes it clear that det(T ) is independent of any basis of M and that
det(ST ) = det(S) · det(T ), for S, T ∈ EndR(M).
In particular we have
T ∈ AutR(M) if and only if det(T ) ∈ R×.
Finally if AT = [aij] is the matrix of T with respect to the basis {mi} of M , then
det(T ) ·m1 ∧ · · · ∧mn = T (m1) ∧ · · · ∧ T (mn)
=
(n∑i=1
ai1mi
)∧ · · · ∧
(n∑i=1
ainmi
)
=
(∑σ∈Sn
sgn(σ)a1σ(1) · · · anσ(n)
)·m1 ∧ · · · ∧mn,
recovering the usual matrix formula for det(T ).
3.3 The trace
Just as we defined the determinant canonically so we can define the trace tr(T ) of an endomorphismT ∈ EndR(M).
Let ∂T : Λn(M)→ Λn(M) be the (well-defined)R-module map given, for any vectors x1, . . . , xn ∈Mby
∂T (x1 ∧ · · · ∧ xn) =n∑j=1
x1 ∧ · · · ∧ T (xj) ∧ · · · ∧ xn,
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where in the jth term T is applied to xj and all other xi are unchanged. Then we define tr(T ) to be thescalar by which ∂T acts on the rank-one R-module Λn(M):
∂T = tr(T ) · id on Λn(M).
This definition of tr(T ) does not require a choice of basis of M . However, if AT = [aij] is the matrixof T with respect to a basis {m1, . . . ,mn} of M then for each j we have
is the sum of the diagonal entries of AT , recovering the usual formula for the trace.
If n = 1 then tr(T ) = det(T ). Hence for general n ≥ 1 we have
tr(Λn(T )) = det(T ).
3.4 Interpolating between trace and determinant
LetM be a freeR-module of rank n and let T ∈ EndR(M). We know explicit formulas for tr(Λ1(T )) =tr(T ) and tr(Λn(T )) = det(T ). In this section we interpolate between these two to give an explicitformula for tr(Λk(T )) for any 1 ≤ k ≤ n.
Choose a basis {mi} of M and let A = [aij] be the matrix of T . Given a k-element subset I = {i1 <· · · < ik} ⊂ [1, n], let
MI = {mi : i ∈ I} and M ′I = {mj : j /∈ I}
so that M = MI ⊕M ′I with corresponding inclusion and projection maps
MIι↪→M
π−→MI .
Define TI to be the composition
TI : MIι↪→M
T−→Mπ−→MI .
For example if k = 1 and I = {i} then TI = aii · id on MI = Rmi.
Let {mI : I ∈ [1, n]k} be the basis of Λk(M) from Prop. 3.2.
Λk(T )mI = T (mi1) ∧ · · · ∧ T (mik)
= TI(mi1) ∧ · · · ∧ TI(mik) +∑J 6=I
aJmJ
= det(TI) ·mI +∑J 6=I
cJmJ
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for some coefficients cJ ∈ R. It follows that
tr(Λk(T )) =∑
I∈[1,n]k
det(TI). (5)
When k = 1 this is tr(T ) and when k = n the sum consists of one term, namely det(T ).
3.5 Vector spaces
Now let F be a field and let V be a vector space over F of dimension dimV = n. We write End(V )for EndF (V ).
Proposition 3.3 A subset {v1, . . . , vk} ⊂ V is linearly independent if and only if v1 ∧ · · · ∧ vk 6= 0.
Proof: If {v1, . . . , vk} ⊂ V is linearly dependent then some vi is a linear combination of the other vj .By multilinearity v1 ∧ · · · ∧ vk is a sum of terms each with repeated factors hence is zero.
If {v1, . . . , vk} ⊂ V is linearly independent then it is contained in a basis {v1, . . . , vk, vk+1, . . . , vn} ofV . By Lemma 3.1 we have
v1 ∧ · · · ∧ vk ∧ vk+1 ∧ · · · ∧ vn 6= 0,
so v1 ∧ · · · ∧ vk 6= 0. �
Proposition 3.4 Let {u1, . . . , uk} and {w1, . . . , wk} be two linearly independent subsets of V , span-ning subspaces U and W respectively, of V . Then U = W if and only if u1∧· · ·∧uk and w1∧· · ·∧wkare proportional in Λk(V ).
Proof: If U = W the map T : U → U given by T (ui) = wi is an isomorphism which induces anisomorphism Λk(T ) : Λk(U)→ Λk(U), such that w1 ∧ · · · ∧ wk = det(T ) · u1 ∧ · · · ∧ uk.
Conversely, supposew1∧· · ·∧wk = c·u1∧· · ·∧uk for some c ∈ F . Then for each i, ui∧w1∧· · ·∧wk =0, so {ui, w1, . . . , wk} is dependent, by Prop. 3.3. Since the wj are independent, it follows that uibelongs to the span of the wj , so ui ∈ W . As i was arbitrary, we have U ⊂ W so U = W .
3.5.1 The characteristic polynomial
An eigenvector for T ∈ End(V ) is a nonzero vector such that T (v) is a scalar multiple of V , sayT (v) = λv. This scalar λ is an eigenvalue for T . Eigenvectors with eigenvalue λ exist if and only ifker(λIV − T ) 6= 0, in other words, when det(λIV − T ) = 0. Therefore the eigenvalues of T are theroots of the characteristic polynomial det(xIV − T ). Here xIV − T is an endomorphism of the free
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F [x]-module F [x]⊗F V and by definition, det(xIV − T ) is the trace of Λn(xIV − T ) on the rank-oneF [x]-module 4
Λn(F [x]⊗F V ) = F [x]⊗F Λn(V ).
Recall that for 1 ≤ k ≤ n we have endomorphism Λk(T ) ∈ End(Λk(V )).
Proposition 3.5 The the expansion of the characteristic polynomial in powers of x is given by
det(xIV − T ) =n∑k=0
(−1)n−k tr(Λk(T )) xn−k.
Proof: It is equivalent to prove that
det(xIV + T ) =n∑k=0
tr(Λk(T )) xn−k.
Choose an F -basis {v1, . . . , vn} of V , so that the n-form ω = v1∧· · ·∧vn generates the one-dimensionalF -vector space Λn(V ). Identifying 1⊗vi with vi, we may also regard {vi} as an F [x]-basis of F [x]⊗FVand ω as an F [x]-module generator of Λn(F [x]⊗F V ). Computing in the latter exterior power, we have
det(xIV + T ) · ω = (xv1 + T (v1)) ∧ · · · ∧ (xvn + T (vn))
= xnω + xn−1
(n∑i=1
v1 ∧ · · · ∧ T (vi) ∧ · · · ∧ vn
)
+ xn−2
( ∑1≤i<j≤n
v1 ∧ · · · ∧ T (vi) ∧ · · · ∧ T (vj) ∧ · · · ∧ vn
)+ · · ·+ det(T ) · ω
Consider a general term corresponding to I = {i1 < · · · < ik}. In the notation of section 3.4 we have
From Prop. 5 it follows that in the rank-one F [x]-module Λn(F (x)⊗F V ) we have
det(xIV + T ) · ω =n∑k=0
xn−k ·∑
I∈[1,n]k
det(TI) · ω =n∑k=0
xn−k · tr(Λk(T ))ω,
so
det(xIV + T ) =n∑k=0
xn−k · tr(Λk(T )),
as claimed. �4Note the left side is an exterior power of an F [x]-module and the right side is an exterior power of an F -module, with
scalars extended to F [x].
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3.5.2 A topological interpretation
Let V be a real vector space of dimension n. A lattice in V is a subgroup L ⊂ V generated by abasis {vi} of V . The quotient X = V/L is a torus (a product of n-copies of the circle S1), whose kth
homology group (with real coefficients)Hk(X) is isomorphic to Λk(V ). Explicitly, for each I ⊂ [1, n]kthe image of VI in X is a subtorus XI of dimension k and the fundamental classes of the XI forI ∈ [1, n]k form a basis of Hk(X). Any continuous mapping f : X → X determines a linear maphk(f) : Hk(X)→ Hk(X). The Lefschetz number of f is defined 5 as the alternating sum of traces:
L(f) =n∑k=0
(−1)n−k tr(hk(f)).
For example if f = id then L(f) is the Euler characteristic (which is zero for X).
Suppose now that f is smooth, with finite fixed-point setXf . At each fixed point p ∈ Xf , the derivativef ′ gives a linear endomorphism f ′p ∈ End(V ) 6 and the Lefschetz fixed-point theorem asserts that
L(f) =∑p∈Xf
sgn[det(I − f ′)p].
Any endomorphism T ∈ End(V ) satisfying T (L) ⊂ L gives a smooth map fT : X → X via fT (v +L) = T (v) + L. Identifying Hk(X) = Λk(V ), we have
hk(fT ) = Λk(T ), so L(fT ) =n∑k=0
(−1)n−k tr(Λk(T )) = det(IV − T ). (6)
The fixed-points of fT are the v + T such that T (v) − v ∈ L. This fixed-point set is finite if and onlyif det(IV − T ) 6= 0 (exercise), in which case it is given by
XfT = (IV − T )−1L/L ' L/(IV − T )L. (7)
The derivative f ′T = T , so det(I − f ′T )p = det(IV − T ) at every fixed-point p. So the Lefschetzfixed-point theorem becomes the equality
L(fT ) = |XfT | · sgn[det(I − T )].
Comparing with (6) and (7), we get the purely algebraic result
| det(IV − L)| = [L : (IV − T )L]. (8)
We will shortly give an algebraic proof of equation (8).
5It is customary to define the Lefschetz number in terms of cohomology. Using Poincare duality it may be phrased interms of homology, which is simpler in our context.
6The tangent spaces of X may be identified with V .
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3.5.3 Grassmannians
The Grassmannian of k-planes is the set Gk(V ) of k-dimensional subspaces of V . This generalizes theprojective space P(V ) = G1(V ) of lines in V . According to Prop. 3.4 we have an injective mapping
λ : Gk(V ) ↪→ P(ΛkV ),
sending a k-plane U ⊂ V to the line [U ] ∈ P(ΛkV ) through the k-fold wedge of any basis of U . Thisis the Plucker embedding of the Grassmannian. As a subset of P(ΛkV ), it turns out that Gk(V ) is anintersection of quadrics, called Plucker relations.
Example: Planes in four dimensions If k = 1 or n − 1 then the Grassmannian is just a projectivespace. The first new case, the one originally considered by Plucker, is k = 2 and dimV = 4. Assume2 6= 0 in F . In this case we have a quadratic form
q : Λ2V −→ Λ4V, q(ω) = 12ω ∧ ω
defining a quadricQ = {[ω] ∈ P(Λ2V ) : q(ω) = 0}.
If ω is a pure wedge ω = u ∧ v, then q(u ∧ v) = 0, so via the embedding u ∧ v 7→ [u ∧ v] ∈ P(Λ2V ),we have
G2(V ) ⊂ Q.
We will show this is an equality.
For any ordered basis (e1, e2, e3, e4) of V we write a nonzero 2-form as
ω =∑i<j
ωijei ∧ ej,
with coefficients ωij ∈ F . We identify Λ4V = F , via the basis element e1234. One checks that
q(ω) = ω12ω34 − ω13ω24 + ω14ω23.
We must show that if q(ω) = 0 then ω = u ∧ v for some u, v ∈ V . Consider the map
ω∧ : V −→ Λ3V, v 7→ ω ∧ v.
Using the ordered bases (e1, e2, e3, e4) and (e234, e134, e124, e123), the matrix of ω∧ is
Mω =
0 ω34 −ω24 ω23
ω34 0 −ω14 ω13
ω24 −ω14 0 ω12
ω23 −ω13 ω12 0
Since ω 6= 0, some diagonal 2 × 2 minor ω2
ij 6= 0. It follows that dim kerMω ≤ 2. Remarkably, the3 × 3 minors of detMω are (−1)i+jωijq(ω), so that detMω = q(ω)2. It follows that q(ω) = 0 if andonly if dim kerMω = 2. In this case we may choose our basis {ei} so that kerMω = {e1, e2}, makingfirst two columns of Mω equal to zero. This implies that ω = ω12e1 ∧ e2. Thus we have proved thatG2(V ) = Q, as claimed.
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3.5.4 Oriented Grassmannians and the characteristic polynomial
We now take F = R, the field of real numbers. Let U be an R-vector space of dimension k. Thenthe set of nonzero k-forms ω ∈ ΛkU is a disjoint union of two open half-lines. An orientation on Uis a choice of one of these half-lines. We write (U, ω) to denote the oriented vector space U whereω ∈ ΛkU belongs to the chosen orientation. An ordered basis (e1, . . . , ek) of U is an oriented basis for(U, ω) if e1 ∧ e2 ∧ · · · ∧ ek belongs to the half-line containing ω. An element g ∈ GL(U) acts on ΛkUby the scalar det g, so any two oriented bases of U differ by transformation g ∈ GL(U) with det g > 0.
Now let V be an R-vector space of dimension n. The oriented Grassmannian Gk(V ) is the set ofpairs (U, ω) where U is a k-plane in V and ω is an orientation on U .
Suppose we also have a positive definite quadratic form q : V → R. The restriction qU of q to anysubspace U ⊂ V is again a positive definite quadratic form on U . Each orientation ω of U may be takento be of the form ω = e1 ∧ · · · ∧ ek where (e1, . . . , ek) is an oriented orthonormal basis of U . Sinceany two such bases differ by a transformation from SO(U, qU), it follows that we have a well-definedembedding
Gk(V ) ↪→ ΛkV, (U, ω) 7→ ω, (9)
which covers the Plucker embedding Gk(V ) ↪→ P(ΛkV ).
If k = 1 this identifies G1(V ) with the unit sphere S = {v ∈ V : |v| = 1}, whereby an oriented line(L, ω) in V corresponds to the point ω ∈ S where the line exits S. This embedding gives a geometricinterpretation of the trace of a transformation T ∈ End(V ), as follows. Associated to q we have thebilinear form 〈u, v〉 = 1
2[q(u+ v)− q(u)− q(v)]. Then we have
tr(T )
n=
∫S
〈Tω, ω〉, (10)
where the integral is taken with respect to the unique SO(V )-invariant measure on S such that∫S
1 = 1.That is, the average of the eigenvalues of T on V is the average over all ω ∈ S of the signed lengths ofthe orthogonal projections of Tω back to the oriented line through ω.
We will prove (10) algebraically, by showing how the integral can be computed by differentiation. Theintegrand is the restriction to S of the new quadratic form qT (v) := 〈Tv, v〉 and
∫S
is an SO(V, q)-invariant linear functional on the space Q of quadratic forms on V . On the other hand, if {ei} is aq-orthonormal basis then the Laplacian ∆ =
∑∂2ei is also an SO(V, q) invariant functional on Q. I
claim that all SO(V, q) invariant functionals on Q are proportional. Granting this, we have∫Sf = c∆f
for every f ∈ Q. Taking f = q, we find c = 1/2n. It follows that∫S
〈Tω, ω〉 =∆qT2n
.
If [tij] is the matrix of T with respect to {ei}, one checks that
qT =∑i,j
tijxixj,
15
where {xi} is the dual basis, and that
∆qT = 2∑i
tii = 2 tr(T ),
and (10) follows. It remains to prove that all SO(V, q)-invariant functionals on Q are proportional.Using q we identify V = V , whence Q = Q, so it is equivalent to prove that all SO(V, q)-invariantelements in Q are proportional. If p is such, then p is constant on S, say p = c on S. Then p− cq = 0on S. Since p and q are homogenous, we have p = cq on V , as desired, so the proof of (10) is complete.
The analogue of (10) for k = n is well-known: det(T ) is the signed volume of Tω, where the sign is+ iff T is orientation-preserving.
The interpolation of these results for 1 ≤ k ≤ n is as follows. The oriented Grassmannians Gk(V ) arecompact submanifolds of ΛkV generalizing the unit sphere for k = 1. The special orthogonal groupSO(V, q) preserves oriented orthonormal bases, hence acts on Gk(V ). In fact this action is transitive,so there is a unique SO(V )-invariant measure assigning unit volume to Gk(V ). Using this measure,we can compute the average of continuous functions on Gk(V ).
Given T ∈ End(V ) and an oriented k-plane ω ∈ Gk(V ), we define a number 〈Tω, ω〉 as follows. IfdimTω < k then 〈Tω, ω〉 = 0. Assume dimTω = k and give Tω the orientation transferred to ω byT . Then 〈Tω, ω〉 is ± the k-dimensional volume of the orthogonal projection of Tω back to ω, wherewe choose + iff T : ω → Tω is orientation preserving. Thus, for any T ∈ End(V ) we have defined afunction ω 7→ 〈Tω, ω〉 on Gk(V ). Then we have(
n
k
)−1· tr(T,ΛkV ) =
∫Gk(V )
〈Tω, ω〉. (11)
That is, the average of the eigenvalues of T on ΛkV is the average of the signed volumes 〈Tω, ω〉 overall oriented k-planes in V . This result can be proved in the same way as (10), by analyzing the actionof SO(V, q) on the space of quadratic forms on ΛkV .
4 Clifford algebras
This is our third quotient of the tensor algebra. Let F be an infinite field. Let V be a vector spaceover F , with dimF (V ) = n < ∞. A quadratic form is a polynomial function q : V → F ofdegree two. That is, q ∈ S2(V ). Let I(V, q) be the ideal in the tensor algebra T (V ) generated by{v ⊗ v − q(v) : v ∈ V }. The Clifford algebra is the F -algebra
C(V, q) := T (V )/I(V, q).
We see immediately that C(V, q) is a deformation of the exterior algebra Λ(V ) = C(V, 0), but more istrue: Since I(V, q) is not homogeneous, the Clifford algebra is filtered but not graded, see section 1.1.From (3) the associated graded algebra of C(V, q) is determined by the ideal σ(I(V, q)) generated byall symbols σ(v ⊗ v − q(v)) = v ⊗ v, we have
σ(I(V, q)) = Ialt(V ),
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so thatgrC(V, q) ' Λ(V ),
as graded algebras. It follows that if {e1, . . . , en} is a basis of V then the products {eI : I ⊂ [1, n]}form an F -basis of C(V, q), where
eI =∏i∈I
ei,
product taken in increasing order of the subscripts, with e∅ = 1.
Though the terms in each generator of I(V, q) do not have the same degree, their degrees do have thesame parity (even). This implies that C(V, q) does have a Z/2Z-grading. Namely, for a ∈ Z/2Z, letCa(V, q) be the projection to C(V, q) of all T i(V ) with i ≡ a mod 2. We then have
Ca(V, q) · Cb(V, q) ⊂ Ca+b(V, q).
Note that C0(V, q) is a subalgebra of C0(V, q), with F -basis {eI : |I| even}, and that C1(V, q) is aC0(V, q)-submodule of C(V, q), with F -basis {eI : |I| odd}.
Associated to q is the bilinear form
b(u, v) = q(u+ v)− q(u)− q(v).
In C(V, q) we have the relation
q(u+ v) = (u+ v)2 = q(u) + uv + vu+ q(v),
hence the relationuv + vu = b(u, v).
In particular, we have uv = −vu if b(u, v) = 0.
4.1 Orthogonal bases and the center of the Clifford algebra
From now on we also assume that 2 6= 0 in F and that q is non-degenerate, meaning that if u ∈ V andb(u, V ) = {0} then u = 0.
Under these conditions, one can show that V has an orthogonal basis {e1, . . . , en}, meaning thatb(ei, ej) = 0 if i 6= j. Non-degeneracy implies that q(ei) 6= 0 for all i. From our remarks above, thevectors {eI : I ⊂ [1, n]} form an F -basis of C(V, q). By induction on I , one checks that
eIeJ = (−1)〈I,J〉eJeI , (12)
where 〈I, J〉 = |I| · |J |+ |I ∩ J |.
Of particular importance is the element
ω := e[1,n] = e1e2 · · · en.
17
One checks thatω2 = D,
where D is the discriminant of the quadratic form q, given by
D = (−1)n(n−1)/2q(e1)q(e2) · · · q(en).
(This is independent of the choice of orthogonal basis.) Thus, inside of C(V, q) we have a quadraticalgebra
F [ω] = {a+ bω : a, b ∈ F} ' F [x]/(x2 −D).
This algebra is closely related to the center of C(V, q), as follows.
Let Z(V, q) be the center of the algebra C(V, q) and let Z0(V, q) be the center of the even subalgebraC0(V, q). From (12) it follows that
Z(V, q) =
{F [ω] for n oddF for n even
, Z0(V, q) =
{F for n oddF [ω] for n even.
(13)
In all cases we haveZ(V, q) ∩ C0(V, q) = F. (14)
An F -algebra A is called central-simple if its center is just F . The Skolem-Noether theorem assertsthat every automorphism of a central simple F -algebra is conjugation by a unit in A×.
Each formula in (13) accords with the Skolem-Noether theorem, as follows. The algebra C(V, q) has aunique automorphism α such that
α(v1 . . . vk) = (−1)kv1 · · · vk for all v1, . . . , vk ∈ V . (15)
When n is even then C(V, q) is central-simple and α is conjugation by ω ∈ C(V, q)×. When n is oddthen C0(V, q) is central-simple and both α and conjugation-by-ω are trivial automorphisms of C0(V, q).
4.2 Orthogonal groups
We have seen that the group SO3 of rotations of three-dimensional Euclidean space has the two-foldcovering S3 → SO3 by the unit sphere S3 in the quaternion algebra H. This generalizes to higherdimensions and general fields F , in the setting of Clifford algebras C(V, q). First we need to prove thatorthogonal groups are generated by reflections.
We continue to assume that 2 6= 0 in F and that q is nondegenerate. The orthogonal group is thelinear isometry group of q:
O(V, q) = {g ∈ GL(V ) : q(gv) = q(v) ∀v ∈ V }.
If v ∈ V and q(v) 6= 0 then V = Fv ⊕ v⊥, where v⊥ = {u ∈ V : b(v, u) = 0} is the hyperplaneorthogonal to v. There is a unique element rv ∈ O(V, q) such that rv(v) = −v and rv(u) = u for allu ∈ v⊥. We call rv the reflection about v⊥. An explicit formula is
rv(x) = x− 〈x, v〉q(v)
v, ∀ x ∈ V.
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Proposition 4.1 The group O(V, q) is generated by reflections. In fact each g ∈ O(V, q) can beexpressed as a product of at most n reflections.
Proof: We only prove the first statement here, since that is all we need. This proof is from Dieudonne“sur les groupes classiques ”. It follows an earlier proof of E. Cartan’s in the real case. We only provethe first statement here.
We argue by induction on n = dimV . If n = 1 then O(V, q) = {±1} and −1 is the unique reflectionon V .
Let g ∈ O(V, q). Since q is nondegenerate there exists v ∈ V with q(v) 6= 0.
Case 1: Suppose gv = v. Then g preserves v⊥. The restriction of q to v⊥ is nondegenerate andthe restriction g of g to v⊥ belongs to O(v⊥, q). By induction, g = r1 · · · rk, for some reflectionsri ∈ O(v⊥, q). For each i define ri ∈ O(V, q) by ri(v) = v and ri = ri on v⊥. Then r1 · · · rk acts on vand v⊥ as g does, so g = r1 · · · rk.
From now on we assume gv 6= v. We consider the vectors v ± gv. Since
q(v ± gv) = 2q(v)± 〈v, gv〉
and q(v) 6= 0, at least one of q(v ± gv) is nonzero.
Case 2: Suppose q(v − gv) 6= 0. Let
u = v − gv, u′ = v + gv.
Then 〈u, u′〉 = 0, and we have ru(u) = −u and ru(u′) = u′. Let x = ru(v) and y = ru(gv). Then
gv − v = −u = ru(u) = x− y and v + gv = u′ = ru(u′) = x+ y.
Solving these equations for x, y we get x = gv, y = v. Thus, ru interchanges v and gv. Hence rugfixes v and we are reduced to Case 1.
Case 3: Suppose q(v+ gv) 6= 0. Letting u = v+ gv, u′ = v− gv and proceeding as in Case 2, we findthat ru(gv) = −v. Hence rvrug fixes v, returning us to Case 1. �
In Case 3 a more intricate argument is needed to control the number of reflections required to expressg. If F = R and q is positive definite, then Case 3 never arises, so in this case the above proof showsthat g is a product of at most n reflections.
4.3 The Clifford group
The Clifford group is defined as
Γ(V, q) = {g ∈ C(V, q)× : gV α(g)−1 ⊂ V },
19
where α is the involution defined in (15). Here we are identifying V with its image in C(V, q), and theproduct
Rg(v) := gvα(g)−1
is taken in C(V, q). Thus, we have a group homomorphism
R : Γ(V, q) −→ GL(V ) g 7→ Rg.
I claim thatkerR = F×.
It is clear that F× ⊂ kerR. Suppose g = g0 + g1 ∈ kerR, with ga ∈ Ca(V, q) for a ∈ Z/2Z. Thismeans for every v ∈ V we have
(g0 + g1)v = v(g0 − g1).
In odd degree, this says g0v = vg0, so g0 ∈ F× by (14). In even degree, this says g1v = −vg1, sog1 = 0 by (12). Hence kerR = F×, as claimed.
I next claim that the image of R lies in O(V, q). For if gvα(g)−1 ∈ V then it is negated by α and wehave
Therefore Rg ∈ O(V, q) for all g ∈ Γ(V, q), as claimed. If q(u) 6= 0 then Ru = ru is the re-flection negating u and fixing the hyperplane u⊥. By Prop. 4.1, the group O(V, q) is generated bysuch reflections ru. It follows that R is surjective, and also that Γ(V, q) is generated, as a group, by{v ∈ V : q(v) 6= 0}. Summarizing, we have exact sequences
1 −→ F× −→ Γ(V, q)R−→ O(V, q) −→ 1
1 −→ F× −→ Γ0(V, q)R−→ SO(V, q) −→ 1,
(16)
where Γ0(V, q) = Γ(V, q)∩C0(V, q)×. Every element of Γ(V, q) may be expressed as a product v1 · · · vk
of vectors vi ∈ V having all q(vi) 6= 0, and Γ0(V, q) consists of such products with k even.
4.4 Clifford conjugation and the spinor norm
Recall that the quaternion algebra H has a conjugation operation x 7→ x, satisfying xy = yx. Anendomorphism of an algebra that reverses the order of multiplication like this is called an antiauto-morphism.
On a general Clifford algebra C(V, q), Clifford conjugation, denoted x 7→ x∗, is the unique antiauto-morphism which is −1 on V . On any product of vectors we have
(v1 · · · vk)∗ = (−1)kvk · · · v1.
20
For example if n = 3, F = R and {e1, e2, e3} is a q-orthonormal basis of R3, then H ' C0(V, q), via
i 7→ e1e2, j 7→ e2e3, k 7→ e3e1,
and each epeq is negated by ∗. Thus, Clifford conjugation on C(V, q) restricts to quaternionic conjuga-tion on C0(V, q), in this case.
Returning to the general case, we define the norm
N : Γ(V, q)→ F×, N(g) = gg∗.
To see that this makes sense, recall that every element g ∈ Γ(V, q) can be written as
g = v1 · · · vk
where all vi ∈ V have q(vi) 6= 0. We have
N(g) = gg∗ = q(v1) · · · q(vk) ∈ F×.
In particular each product gg∗ is central in Γ(V, q). It follows that N is a group homomorphism
N : Γ(V, q)→ F×.
If g ∈ F× then N(g) = g2 ∈ F×2, so N induces a group homomorphism
N : O(V, q) −→ F×/F×2
called the spinor norm. The spinor norm is determined by the formula
N(r1 . . . rk) = q(v1) · · · q(vk),
where each ri = rvi is a reflection for some vi ∈ V with q(vi) 6= 0.
If gg∗ = c2 ∈ F×2 then (g/c)(g/c)∗ = 1. This means the subgroup of Γ(V, q) mapping to F×2 underN is is F× · Pin(V, q), where
Pin(V, q) = {g ∈ Γ(V, q) : N(g) = 1}.
Since F× ∩ Pin(V, q) = {±1}, it follows that we have an exact sequence
For example, if q is a positive definite quadratic form on Rn then N is trivial. In this case, (18) becomesan exact sequence
1 −→ {±1} −→ Spin(V, q)R−→ SO(V, q) −→ 1, (19)
which is the universal covering of SO(V, q), in the sense of topology. For n = 3 we have C0(V, q) = Hand Spin(V, q) = S3; we recover the exact sequence
1 −→ {±1} −→ S3 R−→ SO3 −→ 1,
obtained previously.
Likewise if F is algebraically closed then F× = F×2, so we again obtain (19), which is the universalcovering of SO(V, q), in the sense of algebraic groups. For n = 3 we have C0(V, q) = M2×2(F ) andSpin(V, q) = SL2(F ); we get the exact sequence
1 −→ {±1} −→ SL2(F )R−→ SO3(F ) −→ 1.
For general fields F and quadratic forms q, we have the Witt decomposition of V . This is a direct-sumdecomposition
V = W + V1 +W ′,
where q is identically zero on W and W ′ and is never zero on V1 − {0}. A Witt decomposition isunique up to a transformation from O(V, q). Note that nondegeneracy of q implies that the bilinearform b induces an isomorphism W ′ ' Hom(W,F ). Hence, given a basis {w1, . . . , wr} of W , thereexists a basis {w−1, . . . , w−r} of W ′ such that b(wi, w−j) = 1 if i = j, zero otherwise. Set ωi =wiw−i ∈ C0(V, q). One checks that ω2
i = ωi, that for each t ∈ F× the element
ai(t) := 1 + (t− 1)ωi ∈ C0(V, q)
actually belongs to Spin(V, q), and that ai(t)ai(s) = ai(ts) for all t, s ∈ F×. Thus for each i ∈ [1, r]we have an injective group homomorphism
ai : F× → Spin(V, q).
One further checks that
ai(t) · wj · ai(t)−1 =
{t±1wi if i = ±jwj else.
Thus we have an abelian subgroup of Spin(V, q):
A = 〈ai(t) : i ∈ [1,m], t ∈ F×〉
which acts diagonalizably on V , trivially on V1.
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4.6 Summary
All of the above groups are summarized by the following commutative diagram of maps:
In this diagram the rows are exact sequences and unlabelled arrows are the obvious injections or sur-jections.
5 Lie algebras and their enveloping algebras
Enveloping algebras will be our fourth and last quotients of the tensor algebra. Recall that Cliffordalgebras are deformations of exterior algebras, depending on a quadratic form. We will see that en-veloping algebras are deformations of symmetric algebras, depending on a Lie bracket.
A Lie algebra over a field F is an F -vector space L and a map
L ∧ L −→ L, x ∧ y 7→ [x, y]
such that for all x ∈ L, the endomorphism of L sending y 7→ [x, y] is a derivation.
Explicitly, these conditions mean that the bracket [x, y] is bilinear and alternating and that
[x, [y, z]] = [[x, y], z] + [y, [x, z]]
for all x, y, z ∈ L. This last condition is often written as
[x, [y, z]] + [z, [x, y]] + [y, [z, x] = 0,
and is called the Jacobi identity.
We could have [x, y] = 0 for all x, y ∈ L. Such Lie algebras are called abelian. Every one-dimensionalLie algebra is abelian.
A fundamental example of a Lie algebra is L = End(V ), where V is a vector space over F andthe bracket is given by [X, Y ] = XY − Y X for X, Y ∈ End(V ). More generally, any associativeF -algebra A can be made into a Lie algebra, with bracket [a, b] = ab− ba, for a, b ∈ A.
A homomorphism between Lie algebras L,L′ is a linear mapping φ : L → L′ such that φ([x, y]) =[φ(x), φ(y)] for all x, y ∈ L.
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5.1 Origin and examples: Lie groups
Lie algebras arose classically, and aquired their name, as follows. Let M be a smooth manifold andlet S(M) be the R-algebra of smooth functions f : M → R. A vector field on M is a derivationof S(M). This means X : S(M) → S(M) is a linear endomorphism satisfying the product rule:X(fg) = X(f)g + fX(g) for all f, g ∈ S. A linear combination of vector fields X, Y is again avector field, but the composition XY = X ◦ Y need not be a vector field. However, one can checkthat XY − Y X is a vector field, and that the vector space X(M) of vector fields on M becomes a Liealgebra with bracket [X, Y ] = XY − Y X . If dimM > 0 then X(M) is infinite-dimensional.
Suppose now thatG is a Lie group. This meansG is a smooth manifold with a smooth group structure.We have seen many examples of Lie groups already, such as GL(V ), SL(V ), PGL(V ), SO(V, q) andSpin(V, q) where in all cases V is a finite-dimensional vector space over the fields R or C.
By definition, for each g ∈ G we have a smooth map Lg : G → G given by x 7→ gx. A vector fieldX ∈ X(G) is left-invariant if
X(f ◦ Lg) = (Xf) ◦ Lg, ∀f ∈ S(G).
The subspace Lie(G) ⊂ X(G) of left-invariant vector fields is closed under the bracket, hence is aLie algebra. As a linear space, Lie(G) is canonically isomorphic to Te(G), the tangent space to G atthe identity element e, so Lie(G) is a finite dimensional Lie algebra. Also, any basis of Lie(G) freelygenerates X(G) as an S(G)-module. 7
For the Lie groups above, we have the table
G Lie(G)GL(V ) End(V )SL(V ) sl(V )
PGL(V ) sl(V )SO(V, q) so(V, q)
Spin(V, q) so(V, q)
wheresl(V ) = {X ∈ End(V ) : tr(X) = 0},
so(V, q) = {X ∈ End(V ) : b(Xu, v) + b(u,Xv) = 0 ∀ u, v ∈ V },
where b is the bilinear form associated to q. We note that different Lie groups G,G′ can have Lie(G) 'Lie(G′). This happens precisely when there is a smooth homomorphism G→ G′ with discrete kernel;for such a map induces an isomorphism Te(G)
∼→ Te(G′) on tangent spaces.
5.2 Universal enveloping algebras
We return to an arbitrary field F . Given a Lie algebra L there is a canonical associative F -algebraU(L) containing L such that the bracket in L is given by [x, y] = xy− yx where the products are taken
7For manifolds which are not Lie groups, X(M) need not be free over S(M); one example is the two-sphere M = S2.However, X(M) is always a projective module over S(M).
24
in U(L).
The algebra U(L) is the universal enveloping algebra of L, and is constructed as the quotient
U(L) = T (L)/Ilie(L),
where T (L) be the tensor algebra of the F -vector space L and Ilie(L) is the ideal generated by
{x⊗ y − y ⊗ x− [x, y] : x, y ∈ L}.
If L is abelian then Ilie(L) = Isym(L) is the ideal defining the symmetric algebra Sym(L), so
U(L) = Sym(L)
is just the symmetric algebra of L. For any bracket, the symbol ideal of σ(Ilie(L)) = Isym(L), so wehave
gr(U(L)) = Sym(L),
as graded algebras. This implies that if {x1, . . . , xn} is an ordered basis of L then U(L) has a basisconsisting of monomials
xi11 xi22 · · ·xinn
where each ij ∈ {0, 1, 2, 3, . . . }.8 In particular the inclusion L ⊂ T (L) gives an injection ι : L ↪→U(L), by which we identify L with a Lie subalgebra of U(L), where the latter has the bracket [x, y] =xy − yx, arising from its structure as associative algebra.
The mapping property of U(L) is as follows. If A is an associative F -algebra and
φ : L −→ A
is a Lie algebra homomorphism then there is a unique F -algebra homomorphism φ : U(L)→ A suchthat φ ◦ ι = φ.
5.3 Representations and U(L)-modules
A representation of a Lie algebra L is a Lie algebra homomorphism
ρ : L −→ End(V ),
where V is and F -vector space. By the mapping property for enveloping algebras, ρ extends to give aU(L)-module
ρ : U(L)→ End(V ).
We say the ρ is irreducible if the corresponding U(L)-module is simple.
The image of φ is the subalgebra of End(V ) generated by φ(L), hence the term “enveloping” algebra.The advantage of U(L) over L is that U(L) gives more endomorphisms of V . For example, U(L) canhave a large center, which helps to analyze the structure of the representation (ρ, V ).
8The existence of such a basis is known as the PBW theorem, after H. Poincare, G. Birkhoff and E. Witt.
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Any representation ρ : L → End(V ) gives rise to more representations, via the natural Lie algebrahomomorphisms
End(V ) −→ End(T (V ))
End(V ) −→ End(Sym(V ))
End(V ) −→ End(Λ(V )),
sending X 7→ ∂X , where ∂X is the derivation of T (V ), Sym(V ) or Λ(V ) such that
∂X(v1v2 · · · vk) =k∑i=1
v1v2 · · · vi−1(Xvi)vi+1 · · · vk, (21)
the products being taken in T (V ), Sym(V ) or Λ(V ).
5.4 Representations of sl2
We now consider the most fundamental of all Lie algebras, sl2. This is the Lie algebra with C-basis{e, h, f} and bracket
[h, e] = 2e, [e, f ] = h, [f, h] = 2f. (22)
The enveloping algebra U(sl2) has the basis
{f ihjek : i, j, k = 0, 1, 2, ...}.
To give a representation of sl2, or a module for U(sl2) it is equivalent to give a vector space V overF and three endomorphisms E,H, F ∈ End(V ) satisfying the corresponding relations (22). By themapping property we will then have a homomorphism ρ : U(sl2) → End(V ) such that ρ(f ihjek) =F iHjEk.
We will classify representations ρ : sl2 → End(V ) where dimV is finite. It is a fact, which we shall notprove, that every such representation is a direct sum (in the evident way) of irreducible representations.Therefore it is enough to classify irreducible representations.
The standard representation of sl2 has a two-dimensional vector space V1 with ordered basis {u, v}and operators
E1 =
[0 10 0
], H1 =
[1 00 −1
], F1 =
[0 01 0
],
in terms of the basis. To see this is irreducible, note that any proper submodule would have to be one-dimensional over C, and preserved by h. Since h has distinct eigenvalues ±1 on V1, the submodulewould have be either Cu or Cv. But these lines are not preserved by both e and f .
Now suppose ρ : sl2 → End(V ) is an arbitrary irreducible representation with dimC V finite, and letE = ρ(e), H = ρ(h), F = ρ(f), as above. For every λ ∈ C, let
V (λ) = {v ∈ V : Hv = λv}
26
The relations (22) imply that
EkV (λ) ⊂ V (λ+ 2k), F iV (λ) ⊂ V (λ− 2i).
Since V is finite dimensional and C is algebraically closed, every operator in End(V ) has an eigen-value, so V (λ) 6= 0 for some λ ∈ C. But only finitely many V (λ) can be nonzero, since eigenspacesfor distinct eigenvalues are linearly independent. Hence there is some λ ∈ C with V (λ) 6= 0, butEV (λ) = 0.
Choose a nonzero vectorw ∈ V (λ) again since dimV is finite, there is an integer n such that F nw 6= 0,but F n+1w = 0. Let W be the span of {w, Fw, F 2w, . . . , F nw}. We clearly have FW ⊂ W . Alsoeach F iw belongs to V (λ− 2i), hence is an eigenvector for H , so HW ⊂ W . I claim that EW ⊂ W ,but this is less clear. In U(sl2) the middle relation of (22) can be written as ef = fe+ h. By inductionwe have
ef i = f ie+ f i−1h+ f i−2hf + · · ·+ fhf i−2 + hf i−1,
in U(sl2). It follows that for any i ≥ 1 we have
EF iw = i(λ− i+ 1)F i−1w,
in End(V ). So W is indeed preserved by E, so W = V , by irreducibility. Since F n+1w = 0, this alsoshows that
0 = EF n+1w = (n+ 1)(λ− n)F nw,
and since F nw 6= 0, we must have λ = n. It follows that the actions of E,H, F are given in terms ofthe ordered basis {w,Fw, . . . F nw} by the formulas
EF iw = i(n− i+ 1)F i−1w, HF iw = (n− 2i)F iw, FF iw = F i+1w, (23)
and F n+1w = 0. We have shown that for each integer n ≥ 1 there is at most one simple U(sl2)-moduleVn with dimVn = n + 1. It remains to actually construct such a module. This could be done bychecking that the formulas (23) satisfy the relations (22), but there is a more natural way to constructVn, namely
Vn = Symn V1,
where V1 is the standard representation described above and the operators E1, H1, F1 are extended toVn as in (21). In this picture, the ordered basis {w,Fw, . . . , F nw} becomes {un, un−1v, . . . uvn−1, vn}.Note that we could even take n = 0, in which case V0 = C with e, h, f acting by zero.
To summarize: The Lie algebra sl2 has exactly one irreducible complex representation, up to isomor-phism, in each finite dimension. They can be realized on the symmetric powers Vn of the standardtwo-dimensional representation V1.
Finite dimensional representations of sl2 occur throughout mathematics. We shall see them again inthe classification of conjugacy classes in general linear groups.
27
Part II
Modules over Principal Ideal Domains
In this section we consider the structure of finitely generated R-modules, under the assumption that Ris a Principal Ideal Domain. This will lead to a classification of orbits in Mm×n(R) under GLm(R) ×GLn(R). Taking R = F [X], a polynomial ring over a field F , this will lead to a classification ofconjugacy classes in GLn(F ).
5.5 Modules over Noetherian Rings
In this section R is just a commutative ring with 1. An R-module M is Noetherian if every R-submodule of N is finitely generated.
Proposition 5.1 An R-module M is Noetherian if and only if every ascending chain of submodulesM1 ⊂M2 ⊂ · · · eventually stabilizes.
Proof: The proof is similar to the analogous result for Noetherian rings, and is omitted. �
Lemma 5.2 Let M be an R-module and let N ⊂ M be an R-submodule. Then M is Noetherian ifand only if N and M/N are Noetherian.
Proof: Suppose N and M/N are Noetherian and let M1 ⊂M2 ⊂ · · · be a chain of submodules of M .Let Mi be the image ofMi inM/N . By assumption there is an integer n such thatMi∩N = Mi+1∩Nand Mi = Mi+1 for all i ≥ n. If m ∈ Mi+1 then there is m′ ∈ Mi such that m ≡ m′ mod N . Thismeans m−m′ ∈Mi+1∩N = Mi∩N , so m ∈Mi. Hence Mi = Mi+1 and the chain M1 ⊂M2 ⊂ · · ·stabilizes.
Suppose M is Noetherian. Every chain of submodules of N is also a chain in M , hence must stabilize.Every chain of submodules of M/N is the image of a chain of submodules of M which stabilizes, soany chain in M/N must stabilize. �
Proposition 5.3 If R is Noetherian then then every finitely generated R-module is Noetherian.
Proof: If M is cyclic then M = R/I for some ideal I in R, so the result follows from R beingNoetherian. Using Lemma 5.2, the result follows by induction on the minimum number of generatorsof M . �
Assume thatR is Noetherian and letM be a finitely generatedR-module, with generating set {x1, . . . , xm}.Thus we have a surjective R-module map
whose image is kerφ. Let A ∈ Mm×n be the matrix of ψ with respect to the standard bases of Rn andRm. Thus we have
M ' Rm/ kerφ = Rm/ imψ = Rm/ARn
so the R-module M is completely determined by the matrix A. In fact, for any g ∈ GLm(R) andh ∈ GLn(R) we have
M ' Rm/ARn = Rm/AhRn ' Rm/gAhRn
so the isomorphism class of the R-module M depends only on the orbit of A under the group
Γm×n = GLm(R)×GLn(R)
acting on Mm×n(R) by left and right multiplication. Thus, the classification of R-modules with ngenerators is becomes the problem of classifying the Γm×n orbits on Mm×n(R) for m ≥ 1.
We write A ∼ B if two matrices A,B ∈ Mm×n(R) belong to the same Γm×n-orbit. This means thereis g ∈ GLm(R) and h ∈ GLn(R) such that gAh = B.
5.6 Smith Normal Form: statement and consequences
From now on R is a PID. This means R is a commutative integral domain with identity and every idealin R is generated by just one element.
Theorem 5.4 (Smith Normal Form) If A ∈Mm×n(R) then there is k ≤ m and are nonzero elementsf1, . . . , fk of R, where f1 | f2 | · · · | fk, such that
A ∼
f1 0 0 . . . 00 f2 0 . . . 0
0 0. . .
... fk 0...
0 0. . .
0 0
.
Moreover r is unique and the elements fi are unique up to multiplication by units in R.
The elements f1, . . . , fk are the invariant factors of A.
In terms of modules, the theorem asserts that ifN is a submodule ofRm then there is a basis v1, . . . , vmofRm and a unique sequence f1 | f2 | · · · | fk ∈ R (for some unique k ≤ m) such that {f1v1, . . . , fkvk}is a basis of N .
29
Corollary 5.5 If R is a PID and M is an R-module generated by m elements. Then
1. There is a unique chain of ideals I1 ⊃ I2 ⊃ · · · ⊃ Im of R such that
M ' R/I1 ⊕R/I2 ⊕ · · · ⊕R/Im.
2. The torsion submodule of M is
Mtor ' R/I1 ⊕R/I2 ⊕ · · · ⊕R/Ik,
where k = max{j : Ij 6= 0}; hence M 'Mtor ⊕Rm−k.
3. The minimal number of generators of M is m− `, where ` = max{j : Ij 6= R}.
4. The annihilator of Mtor is the ideal Ik.
Proof: We have seen that M ' Rm/ARn for some integer n and matrix A ∈ Mm×n(R). By 5.4, wehave M ' Rn/BRm where B is the Smith Normal Form of A. Taking Ij = Rfj , items 1, 3 and 4 areimmediate. Item 3 will be proved in section 5.9. �
The number r = m− k is called the rank of M . It is intrinsically determined as follows. Let F be thefraction field of R. Since F ⊗RMtor = 0, the rank r is the dimension of the F -vector space F ⊗RM ,hence is uniquely determined by M .
Corollary 5.6 Let V be a finite dimensional real vector space and letL be a lattice in V . IfA ∈ GL(V )preserves L then [L : AL] = | det(A)|.
Proof: With respect to a basis v1, . . . , vn of L we have A ∈ Mn×n(Z). By Thm. 5.4, there areg, h ∈ GLn(Z) such that gAh is a diagonal matrix with diagonal entries (f1, . . . , fn),where the fi arethe invariant factors of the Z-module L. We have
Applying this result to IV − A gives an algebraic proof of equation (8) above.
5.7 Orbits of SLn(R) on Rn
In this section we prove the m = 1 case of Thm. 5.4. Recall that R is a PID. Given a vector v =(v1, . . . , vn) ∈ Rn we let (v) denote the ideal in R generated by {v1, . . . , vn). Since R is a PID we have(v) = Rd for some d ∈ R.
30
Proposition 5.7 Assume that n ≥ 2. Then for u, v ∈ Rn the following are equivalent:
1. u, v belong to the same SLn(R)-orbit;
2. u, v belong to the same GLn(R)-orbit;
3. (u) = (v).
Proof: The implication (1⇒ 2) is clear.
(2 ⇒ 3): If u = gv for some g ∈ GLn(R) then the components of gv are R-linear combinations ofthe components of v, so (gv) ⊂ (v). Replacing g by g−1 shows the other containment, so we have(gv) = (v).
(3⇒ 1): It suffices to show that any v ∈ Rn lies in the orbit of (d, 0, . . . , 0), whereRd = (v). Dividingeach component of v by d, we may assume d = 1. The problem now is to show that if (v) = R thenthere is a matrix in SLn(R) whose first column is v.
Suppose n = 2, and let v = (v1, v2) have (v) = R. Choose x1, x2 such that x1v1 + x2v2 = 1. Then thematrix [
v1 −x2v2 x1
]belongs to SL2(R), proving the result for n = 2. We now argue by induction. 9
Assume the result holds for n ≥ 2, and let v ∈ Rn+1 have (v) = R. Set v′ = (v1, . . . , vn) and let(v′) = Rd. Then v′ = du where u = (u1, . . . , un) ∈ Rn and (u) = R. By induction, there exists amatrix h ∈ SLn(R) whose first column is u:
h = [u h′],
where h′ ∈ Mn×n−1(R). Since (d, vn+1) = (v) = R there are s and t in R such that td − svn+1 = 1.Set
g =
[v′ h′ suvn+1 0 t
]=
[du h′ suvn+1 0 t
].
Here v′ = du and su have size (n − 1) × 1, and 0 is 1 × (n − 2), so g is an n × n matrix whose firstcolumn is v. We compute
Hence g is a matrix in SLn+1(R) whose first column is v. �
Note that the same result holds with the first column replaced by any column. Taking transposes, theresult also holds for any row. We have proved Thm. 5.4 for m = 1.
9See E. Schenkman, The basis theorem for finitely generated abelian groups Amer. Math. Monthly (1960) 770-771.
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5.8 Existence of Smith Normal Form
In this section let A ∼ B means that A,B ∈Mm×n(R) belong to the same SLm(R)× SLn(R)-orbit.
Lemma 5.8 Assume m ≥ 2. If A ∈Mm×n(R) then
A ∼[b 0′
0′′ B
],
where (row1(A), col1(A)) ⊂ Rb, B ∈Mm−1,n−1(R), 0′ is a row vector in Rn−1, 0′′ is a column vectorin Rm−1, and the ideal (row1(A), col1(A)) ⊂ Rb.
Proof: In this proof all matrices will be partitioned as in the statement of the lemma and ∗, ∗′, ∗′′ willdenote variable matrices over R of the correct sizes.
By induction on m, we may assume row1(A) is nonzero. Applying a permutation matrix h0 ∈ SLn(R)(changing one nonzero entry to −1 if necessary to make det(h0) = 1) we have the matrix
A0 := Ah0 =
[a0 ∗′∗′′ ∗
], a0 6= 0.
We now apply Prop. 5.7 repeatedly:
There is g1 ∈ SLm(R) such that
A1 := g1A0 =
[a1 ∗′0′′ ∗
], with Ra0 ⊂ Ra1 = (col1(A0)).
Next there is h2 ∈ SLn(R) such that
A2 := A1h2 =
[a2 0′
∗′′ ∗
], with Ra1 ⊂ Ra2 = (row1(A1)).
Next there is g2 ∈ SLm(R) such that
A3 := g3A2 =
[a3 ∗′0′′ ∗
], with Ra2 ⊂ Ra3 = (col1(A2)).
Repeating this over and over we get a chain of ideals:
Ra0 ⊂ Ra1 ⊂ Ra2 ⊂ · · · ⊂ R
which must stabilize because R is Noetherian. Thus, there is i ≥ 0 such that ai+1 = uai for some unitu ∈ R×.
Suppose i is odd. Then
Ai =
[ai ∗′0′′ ∗
],
32
where now ai divides each entry of ∗′. The matrix h =
[1 − ∗′ /ai0′′ I
]lies in SLn(R) and Aih is of
the form asserted in the lemma.
If i is even then
Ai =
[ai 0′
∗′′ ∗
],
where ai divides each entry of ∗′′. The matrix g =
[1 0′
− ∗′′ /ai I
]lies in SLm(R) and gAi is of the
form asserted in the lemma.
The lemma is proved. �
Lemma 5.9 Suppose we have a matrix[b 0′
0′′ B
]as in Lemma 5.8, such that some entry of B is not
divisible by b. Then [b 0′
0′′ B
]∼[c 0′
0′′ C
],
where Rb ( Rc.
Proof: Suppose v = rowi(B) is a row of B containing an entry not divisible by b. Let g ∈ SLm(R)be the matrix sending the standard basis vector ei 7→ e1 + ei and fixing all the other ej’s. Then
g
[b 0′
0′′ B
]=
[b v0′′ B
].
By Lemma 5.8 we have [b v0′′ B
]∼[c 0′
0′′ C
],
where Rb and (v) are contained in Rc. Since (v) is not contained in Rb we have Rb ( Rb+ (v) ⊂ Rc.�
Now let A ∈Mm×n(R). Applying Lemmas 5.8 and 5.9 alternately, we have
A ∼[b 0′
0′′ B
]∼[c 0′
0′′ C
]∼[d 0′
0′′ D
]∼ . . .
with proper containments of ideals Rb ( Rc ( Rd . . . .
Since R is Noetherian, eventually the situation of Lemma 5.9 cannot occur. Thus we arrive at
A ∼[f1 0′
0′′ F1
],
where f1 is a linear combination of the entries of A and f1 divides every entry of the matrix F1.Repeating all of this with A replaced by F1 we have
F1 ∼[f2 0′
0′′ F2
],
33
where f2 divides every entry of F2 and is a linear combination of the entries of F1, so f1 | f2.
Repeating this process we arrive at the invariant factors f1 | f2 | · · · | fk and we have proved theexistence of Smith Normal form.
5.9 Uniqueness of Smith Normal Form
The aim of this section is to prove that each SLm(R) × SLn(R)-orbit in Mm×n has exactly one set ofinvariant factors f1 | f2 | · · · | fk. In fact we will show how the fi may be computed directly from A.This eliminates the need for row and column operations, which are fine for proving existence, but quiteboring to implement in practice.
Let A ∈ Mm×n(R). Choose a set of j columns and j rows of A and take the determinant of theresulting j× j matrix. This determinant is a j-minor of A. From our discussion of exterior powers, wesee that the j-minors of A are the matrix entries of the exterior power matrix Λj(A) ∈ End (Λj(Rn)).Let Ij(A) be the ideal in A generated by all j-minors of A. We note that
Ij(A) = I1(Λj(A)).
Since each minor is an R-linear combination of j − 1-minors, we have
I1(A) ⊃2 (A) ⊃ · · · ⊃ Iρ(A),
where ρ = ρ(A) is the largest value of j such that Ij(A) 6= 0. Equivalently, ρ(A) is the largest value ofj such that Λj(A) is a nonzero matrix. 10
Lemma 5.10 If A ∼ B ∈Mm×n(R) then Ij(A) = Ij(B) for all j.
Proof: Writing B = gAh we clearly have I1(A) ⊃ I1(B). By symmetry we have I1(A) = I1(B).Now Λj(B) = Λj(g)Λj(A)Λj(h), so
Ij(A) = I1(Λj(A)) = I1(Λ
j(B)) = Ij(B).
�
Now if
A ∼
f1 0 0 . . . 00 f2 0 . . . 0
0 0. . .
...... fk 0
...0 0
. . .0 0 . . . 0
10We can also characterize ρ(A) as the rank of the image ARn. There is an unfortunate conflict of terminology here,
because we have been speaking of the “rank” r of Rm/ARn. The two notions of rank are related by: ρ(A) + r = m.
34
with f1 | f2 | · · · | fk then Rf1 · · · fj = Ij(A) for each j, so each fj is uniquely determined, up toa unit in R, by the minors of A. This proves uniqueness of the Smith Normal Form, as well as theconverse to Lemma 5.10:
Corollary 5.11 We have A ∼ B ∈Mm×n(R) if and only if Ij(A) = Ij(B) for all j.
5.10 Summary and variations
When a group G acts on a set X with many orbits, one tries to classify the orbits in two steps:
i) Decompose X =⊔ι∈I Xι into G-orbits Xι, where ι ranges over some index set I.
ii) Find functions φj : X → I, 1 ≤ j ≤ m whose values on any given x ∈ X determine the G-orbitXι containing x. These functions are the basic invariants for the G-orbits.
Example 1: The group SOn(R) acts on Rn. The orbit decomposition is
Rn =⊔r≥0
Sr
where Sr of radius r ≥ 0. We need only one basic invariant: the quadratic form q(x) =∑x2i , and
Sr = {x ∈ Rn : q(x) = r2}.
Example 2: Let R be a PID. The group G = SLn(R) acts on Rn. Again we need only one basicinvariant, which assigns to v ∈ Rn the ideal (v) ⊂ R generated by the components of v. The orbits areparametrized by the set I of ideals of R and
Rn =⊔I∈I
RnI ,
where RnI = {v ∈ Rn : (v) = I}.
Example 3: R is again a PID. Now G = SLm(R) × SLn(R), acting on Mm×n(R), and the orbitdecomposition is
Mm×n(R) =⊔I∈I
Mm×n(R)I ,
where I is the set of chains of ideals I = (I1 ⊃ I2 ⊃ · · · ⊃ Im), and A ∈Mm×n(R)I iff each invariantfactor fj of A generates Ij for 1 ≤ j ≤ m. The ideals Rf1, . . . , Rfm are the basic invariants of A, andthey can be found by computing the minors of A.
6 Rational canonical form and regular transformations
Let V be a vector space of dimension n over a field F . In this section we classify the orbits of GL(V )acting on EndF (V ), by conjugation. This will follow from Cor. 5.5, applied to the polynomial ring
35
F [X]. Each ideal in I ⊂ F [X] has a canonical generator, namely the unique monic polynomial f ∈ Iof minimal degree.
For each T ∈ EndF (V ) we have a homomorphism
ϕT : F [X] −→ EndF (V ) f(X) 7→ f(T ) (24)
making V into an F [X]-module which we denote by VT . The minimal polynomial of T is the canoni-cal generator of the ideal kerϕT . Since dim(V ) <∞ the F [X]-module VT has rank zero and coincideswith its torsion submodule: VT = (VT )tor.
One checks that for T, S ∈ EndF (V ) we have T ∼ S under GL(V ) if and only if VT ' VS as F [X]-modules. Therefore the classification of GL(V )-orbits in EndF (V ) is equivalent to the classificationof torsion F [X]-modules.
By Cor. 5.5 there is an integer s ∈ [1, n] and unique ideals I1 ⊃ I2 ⊃ · · · ⊃ Is 6= 0 such that
VT 's⊕j=1
F [X]/Ij. (25)
Under this isomorphism the action of T on VT corresponds to multiplication by X on each summandF [X]/Ij .
6.1 Cyclic modules and companion matrices
To make (25) more explicit, we first study the case of one summand:
VT ' F [X]/I,
where I is a nonzero ideal in F [X]. These are precisely the torsion F [X]-modules which are cyclic,meaning they are generated by one element.
The ideal I is also a cyclic F [X] module, and it has a canonical generator, namely the monic polynomialf ∈ I of minimal degree. We write
f = Xd − a1Xd−1 − a2Xd−2 − · · · − ad−1X − ad.
Note that f is the minimal polynomial of the endomorphism Tf ∈ EndF (F [X]/I) given by multipli-cation by X .
As an F -vector space, F [X]/I has the ordered basis {x0, x, . . . , xd−1}, where x is the image of X inF [X]/I , and x0 = 1 + I . With respect to this basis, the matrix of Tf is given by
Af =
0 0 0 . . . ad1 0 0 . . . ad−1
0 1 0. . . ...
... . . . 0 a20 0 . . . 1 a1
,
36
since in F [X]/I we have the relation
xd = a1xd−1 + a2x
d−2 + · · · ad−1x+ a0.
We call Af the companion matrix of f . Using Af one checks that f is also the characteristic polyno-mial of Tf and that the invariant factors of Tf are 1, 1, . . . , 1, f .
6.2 Rational canonical form
Returning to (25), let fj be the canonical generator of Ij . Then we have
f1 | f2 | · · · | fs,s∑j=1
deg(fj) = n,
and fs(x) is the minimal polynomial of T ; this is the monic polynomial generating the annihilator ofVT , that is, the kernel of the map (24).
The matrix AT of T on V with respect to the canonical basis in each summand is a diagonal blockmatrix of companion matrices
.This matrix AT is the rational canonical form of T . In summary, we have
Proposition 6.1 For two endomorphisms S, T ∈ EndF (V ) the following are equivalent.
1. S is conjugate to T by an element of GL(V ).
2. VS ' VT as F [X]-modules.
3. AS = AT .
6.3 Finding the rational canonical form
In this section we show how to find the rational canonical form of any transformation T ∈ EndF (V ).
Proposition 6.2 The rational canonical form of T ∈ EndF (V ) is the block matrix⊕jAfj , where the fjare the nonconstant invariant factors of the transformation TX := X ·IV −T ∈ EndF [X](F [X]⊗F V ).
37
Proof: By reverse-engineering. Suppose f ∈ F [x] is a monic polynomial of degree n > 0 withcompanion matrix Af . Consider the matrix
X · IV − Af ∈Mn(F [X]).
For j < n this matrix has a j-minor equal to 1 and det(X ·IV −Af ) = f . Hence the invariant factors ofX ·IV −Af are 1, . . . , 1, f . This shows that f is the unique nonconstant invariant factor ofX ·IV −Af .
Now suppose T ∈ EndF (V ) has rational canonical form AT = ⊕Afj . Then the matrix of TX in thesame basis is the block matrix
In particular the characteristic polynomial PT (X) = det(TX) is the product of the invariant factors:
PT (X) =s∏j=1
fj.
It follows that the minimal polynomial fs divides the characteristic polynomial PT (X). This provesthe Cayley-Hamilton theorem:
PT (T ) = 0.
As an example, we compute the rational canonical form of the matrix
T =
1 2 34 5 67 8 9
.The matrix
TX =
x− 1 −2 −3−4 x− 5 −6−7 −8 x− 9
has d1 = d2 = 1 and d3 = det(xI − T ) = x3 − 15x2 − 18x is the unique invariant factor. Hence therational canonical form of T is the matrix
AT =
0 0 01 0 180 1 15
.It was not obvious at the outset that T would be a single companion matrix with respect to some basis.In the next section we will characterize such transformations.
38
6.4 Centralizers and regular transformations
Given T ∈ EndF (V ), we consider the centralizer
Z(T ) = {S ∈ EndF (V ) : ST = TS}.
This is an F -subalgebra of EndF (V ). Equivalently,
Z(T ) = EndF [X](VT )
is the algebra of F [X]-module endomorphisms of VT .
The isomorphism (25) expresses VT as direct sum of F [X]-modules F [X]/(fj), where the fj are themonic invariant factors of T . Each of these summands is actually a commutative F -algebra which alsocommutes with the action of T on all of VT . Hence we have
s∏j=1
F [X]/(fj) ⊂ Z[T ],
as F -algebras. In particular dimF Z[T ] ≥∑
j deg fj = n. We say T is regular if dimF Z(T ) = n.
Proposition 6.3 For T ∈ EndF (V ) the following are equivalent.
1. T is regular;
2. TX has just one nonconstant invariant factor;
3. VT is a cyclic F [X]-module;
4. The minimal and characteristic polynomials of T coincide;
5. There is a basis of V for which the matrix of T has the form0 0 0 . . . an1 0 0 . . . an−1
0 1 0. . . ...
... . . . 0 a20 0 . . . 1 a1
,
for some elements a1, . . . , an ∈ F .
Proof: We prove 1⇔ 2; the rest is easily seen from our discussion so far.
If there is a pair nonconstant invariant factors then fs−1, fs would be one such pair. Since fs−1 | fs,we have an evident surjective map of R-modules
F [X]/(fs) −→ F [X]/(fs−1).
39
Extending this map by zero on the summands F [X]/(fj) for j < s, we get an endomorphism inEndF [X](VT ) which does not preserve each summand, so cannot lie in
∏j F [X]/(fj). It follows that
dimZ(T ) > n so T is not regular.
For any commutative ring R with identity and ideal I ⊂ R, we have EndR(R/I) = R/I . Moreprecisely, the endomorphism ring of the R-module R/I is iso the R-algebra R/I . Now if VT only onenonconstant invariant factor f then VT ' F [X]/(f), so
Z(T ) = EndF [X](VT ) = F [X]/(f).
Since deg f = dimV = n it follows that T is regular. �
The set EndF (V )reg of regular transformations in EndF (V ) is a union of GL(V )-orbits under conjuga-tion. In part 5 of Prop. 6.3, the numbers a1, . . . , an are the coefficients of the characteristic polynomialof T ∈ EndF (V )reg, hence the ak are obtained directly from T via
ak = (−1)k−1 tr(ΛkT ),
where tr(ΛkT ) is the trace of the natural map induced by T on the exterior power ΛkV . It follows thatwe have a bijection
The inverse bijection is obtained by choosing an ordered basis (e1, . . . , en) of V so as to identifyEndF (V ) = Mn(F ) and GL(V ) = GLn(F ), and sending (a1, . . . , an) ∈ F n to the the correspondingcompanion matrix in Mn(F ). Passing to GLn(F )-orbits makes this independent of the choice of basis.
The companion matrices form an affine subspace A ⊂ Mn(F ) which meets each regular orbit ofGLn(F ) in Mn(F ) in exactly one point, namely the companion matrix of the orbit. This affine spaceA has a canonical basepoint, namely
Jn =
0 0 0 . . . 01 0 0 . . . 0
0 1 0. . . ...
... . . . 0 00 0 . . . 1 0
(26)
which is the unique matrix in A with minimal(=characteristic) polynomial equal to Xn.
6.5 Conjugacy vs stable conjugacy
.
Corollary 6.4 LetK be a field containing F and suppose S, T ∈ End(V ) are conjugate by an elementof GL(K ⊗F V ). Then S and T are conjugate by an element of GL(V ).
40
Proof: Let f1, . . . , fr be the invariant factors of T , so that
VT 'r⊕i=1
F [x]/(fi).
Since
(K ⊗ V )T ' K ⊗ VT 'r⊕i=1
K[x]/(fi),
the matrix AT with respect to the canonical basis is the same on V and on K ⊗ V . Since the sameholds for S which is conjugate to T by an element of GL(K ⊗ V ), it follows that AS = AT , so S isconjugate to T by an element of GL(V ), by Prop. 6.1. �
Example: Let F = R and consider the two matrices S =
[1 10 1
]and T =
[1 −10 1
]in SL2(R). Then
S is conjugate to T by[i 00 −i
]∈ SL2(C) and by
[1 00 −1
]∈ GL2(R), but S and T are not conjugate
in SL2(R), as you can easily check. We say that S and T are stably conjugate but not rationallyconjugate in SL2(R). Cor. 6.4 shows this phenomenon does not occur in GLn(F ): there stable andrational conjugacy coincide.
7 Jordan Canonical Form
We have seen that Rational Canonical Form of an endomorphism T ∈ End(V ) is based on the invariantfactors of the F [x]-module VT , and this canonical form is unchanged if we extend scalars to a largerfield.
However, the structure of the summands F [X]/(fi) does depend on F . Indeed, take a monic poly-nomial f ∈ F [x] of positive degree, with factorization f = pm1
1 · · · pmkk where p1, . . . , pk are distinct
irreducible monic polynomials in F [x]. By the Chinese Remainder Theorem, we have
F [x]/(f) =k⊕i=1
F [x]/(pmii ),
so the structure of this F [x]-module depends on the factorization of f in F [x]. The simplest situationis when F is large enough to contain all of the roots of f . If f = (x − λ1)m1 · · · (x − λk)mk whereλ1, . . . , λk ∈ F are the distinct roots of f then
F [x]/(f) =k⊕i=1
F [x]/(x− λi)mi .
As a basis of F [x]/(x− λ)m we may take
{1, x− λ, (x− λ)2, . . . , (x− λ)m−1}
41
the matrix of multiplication by x − λ is then all zeros except 1′s below the main diagonal. It followsthat the matrix of X is the Jordan matrix
Jm(λ) =
λ 0 0 . . . 01 λ 0 . . . 0
0 1 λ. . . ...
... 1. . . 0. . . λ 0
0 0 . . . 1 λ
= λI + Jm, (27)
where Jm is the matrix from (26).
Now let f1 | f2 | · · · | fr be the invariant factors of T ∈ End(V ), and suppose all the roots of fr belongto the field F . Let mi(λ) be the multiplicity of λ as a root of fi. Then
F [x]/(f) =⊕λ
r⊕i=1
F [x]/(x− λ)mi(λ),
where λ runs over the roots of fr (that is, the eigenvalues of T ). So there is a basis of V putting thematrix of T in the form
JT =⊕λ
JT (λ), (28)
Where
JT (λ) =r⊕i=1
Jmi(λ)(λ)
is a direct sum of Jordan matrices (27). The decomposition (28) is the Jordan canonical form of T .For each eigenvalue λ of T we have
m1(λ) +m2(λ) + · · ·+mr(λ) = m(λ),
where m(λ) is the multiplicity of λ as a root of det(xI − T ). Since f1 | f2 | · · · | fr, we have
m1(λ) ≤ m2(λ) ≤ · · · ≤ mr(λ);
such a sequence is a partition of m(λ).
Thus, the sequencesµT (λ) = (m1(λ), . . . ,mr(λ))
(for each eigenvalue λ of T ) determine the conjugacy class of T . More precisely:
Proposition 7.1 Let S, T ∈ End(V ) and let K ⊃ F be a field containing the eigenvalues of S and T .Then S and T are conjugate by an element of GL(V ) if and only if µT (λ) = µS(λ) for all λ ∈ K.
Proof: The partitions µT (λ) and µS(λ) determine the invariant factors of S and T in VK , so S and Tare conjugate by an element of GL(K ⊗F V ). By Cor. 6.4, it follows that S and T are conjugate by anelement of GL(V ). �
42
7.1 Semisimple and nilpotent endomorphisms
An endomorphism N ∈ End(T ) is nilpotent if some power of N is zero.
Proposition 7.2 For N ∈ End(T ) the following are equivalent.
1. N is nilpotent.
2. The eigenvalues of N are all zero.
3. tr(Λk(N)) = 0 for all 1 ≤ k ≤ n.
Proof: SupposeN r = 0, and λ is an eigenvalue ofN in some fieldK containing F . There is a nonzerovector v such that Nv = λv. Then N rv = λrv, so λr = 0, which implies that λ = 0. Conversely, if theeigenvalues of N are then all zero, so N is conjugate to a sum
Jn1(0) + · · ·+ Jnr(0)
of strictly upper-triangular Jordan matrices, which is nilpotent. Finally, having all eigenvalues zeromeans that det(xI −N) = xn, so 2 and 3 are equivalent because ± tr(Λk(N)) is the coefficient of xk
in det(xI −N). �
Remark: If F = R or C, there is also a topological condition for nilpotency: N is nilpotent if and onlyif the conjugacy class of N contains 0 in its closure in End(V ). In fact this holds for any algebraicallyclosed field F with the Zariski topology on End(V ).
The invariant factors of a nilpotent element N ∈ End(V ) are xm1 , xm2 , . . . , xmr , where m1 ≤ m2 ≤· · · ≤ mr is a partition of n, and the Jordan form N is
JN =r⊕i=1
Jmi(0).
In particular we have
Proposition 7.3 The set of nilpotent conjugacy classes in End(V ) is in bijection with the set of ofpartitions of n = dimV . In particular there are only finitely many nilpotent classes.
At the other extreme we say that an endomorphism S ∈ End(V ) is semisimple if VS is a semisimpleF [X]-module, meaning that VS is a direct sum of simple F [X]-modules. Now the submodules of theF [X]-module F [X]/(f) are just the ideals in the ring F [X]/(f). It follows that F [X]/(f) is simple ifand only if f is irreducible in F [X].
In the following result we assume F is a perfect field. These are the fields F for which irreduciblepolynomials are separable, meaning they have distinct roots in an algebraic closure of F . Perfect fieldsinclude all fields of characteristic zero, as well as algebraic extensions of finite fields. 11
11The simplest example of an imperfect field is Fp(t), the rational functions in one variable over Fp. Here the polynomialf = Xp − t is irreducible, by Eisenstein’s criterion, but f = (X − τ)p over the field K = F (τ), where τ is a root of f .
43
Proposition 7.4 Assume F is a perfect field. For S ∈ EndF (V ) the following are equivalent.
1. S is semisimple;
2. The minimal polynomial of S is separable;
3. If K ⊃ F is any field containing all the roots of det(xI − S) then K ⊗F V is a direct sum ofeigenlines for S.
Proof: (1⇔ 2): Let f1 | · · · | fk be the invariant factors of S, so that fk is the minimal polynomial ofS and
VS =k⊕i=1
F [x]/(fi).
Factoring each fi, the Chinese Remainder Theorem shows that VS is a direct sum of F [X]-modules ofthe form F [X]/(ge), where g ∈ F [X] is irreducible and e ≥ 1. Now F [X]/(ge) has a unique simplesubmodule, generated by the image of ge−1, hence can only be semisimple if e = 1, in which case itis simple. It follows that each invariant factor fi factors into distinct irreducible polynomials, whichare separable since F is perfect. In particular the minimal polynomial fk is separable. Converselyif fk is separable, then each invariant factor fi, which divides fk, is also separable. Factoring fi intoirreducibles, we see that VS is a direct sum of simple F [X]-modules, so S is semisimple.
(2⇔ 3): We have remarked that if fk is separable then so is every invariant factor fi. Let {λij} be theroots of fi inK. Then (K⊗F V )S is a direct sum of modulesK[x]/(x−λij) which are one-dimensionalover K, and are eigenlines for S. Conversely, if 2 fails then some irreducible factor g of fr appearswith multiplicity e > 1. It follows the Jordan form of S contains a Jordan matrix Je(λ), which has onlyone eigenline, so 3 fails. �
Remark: If F = R or C, there is also a topological condition for semisimplicity: S is semisimple ifand only if the conjugacy class of S is closed in End(V ). Again, this holds for any algebraically closedfield F with the Zariski topology on End(V ).
Proposition 7.5 Let S, S ′ ∈ End(V ) be semisimple. Then S and S ′ are conjugate under GL(V ) if andonly if det(xI − S) = det(xI − S ′).
Proof: Assume that det(xI − S) = det(xI − S ′). Let K be a field containing the eigenvaluesλ1, . . . , λs of S and S ′ and let mj be the multiplicity of λj . By Prop 7.4, S and S ′ are both conjugatevia GL(K ⊗ FV ) to the diagonal matrix
(λ1, . . . , λ1︸ ︷︷ ︸m1
, λ2, . . . , λ2︸ ︷︷ ︸m2
, . . . , λs, . . . , λs)︸ ︷︷ ︸ms
.
From Cor. 6.4 we conclude that S, S ′ are conjugate via GL(V ). �
44
7.2 The Jordan decomposition
A Jordan matrix Jm(λ) can be written as the sum
Jm(λ) = λI + Jm(0),
a scalar matrix λI plus a nilpotent matrix Jm(0), and these two matrices commute. It follows that theJordan form has a basis-free interpretation:
Proposition 7.6 (Jordan decomposition) Every endomorphism T ∈ End(V ) may be uniquely writtenas T = S +N , where S is semisimple, N is nilpotent, and SN = NS.
The last condition means that N belongs to the centralizer
where S ranges over the semisimple endomorphisms in End(V ) and EndS(V )nil is the set of nilpotentendomorphisms of EndS(V ). Note that S is the unique semisimple element in S + EndS(V )nil.
If S is a given semisimple conjugacy class in End(V ) then the set
End(V )S :=∐S∈S
[S + EndS(V )nil]
consists of finitely many conjugacy classes, namely those those with a given characteristic polynomial.
In geometric terms, we have a mapping
χ : End(V ) −→ F n, χ(T ) = (tr(T ), tr(Λ2T ), . . . , tr(ΛnT )),
which assigns to T ∈ End(V ) the vector of coefficients (up to signs) in the characteristic polynomialdet(xI − T ). The sets End(V )S are the fibers of χ. If F = R or C or is algebraically closed, then Sis the unique closed conjugacy class in End(V )S . Moreover End(V )S is itself the closure of a uniqueconjugacy class, namely, that of the companion matrix Cf , where f is the characteristic polynomial forS.
7.3 Nilpotent elements and sl2
Now assume the ground field F = C (or any algebraically closed field of characteristic zero).
For each partition n = m1 +m2 + · · ·+mr, with 1 ≤ m1 ≤ · · · ≤ mr, we also have a representation
V = Vm1−1 ⊕ Vm2−1 ⊕ · · · ⊕ Vmr−1
45
of sl2, where as in section 5.4, Vd = Symd V1 and V1 is the standard two-dimensional representation. Ifd ≥ 1 the representation sl2 → End(Vd) is injective. It follows that the representation sl2 → End(V )is injective if mr ≥ 2. In each summand, the element f ∈ sl2 acts via the Jordan matrix Jmk
, inthe f -cyclic basis. Hence each nonzero nilpotent element N ∈ End(V ) is the image of f under aninjective representation ρN : sl2 → End(V ). Moreover the conjugacy class of N and the isomorphismclass of ρN determine each other. Thus we have bijections
End(V )nil/GL(V ) Irrn(sl2)
N ⇀ ρN
ρ(f) ↽ ρ
between the set of GL(V )-conjugacy classes of nilpotent endomorphisms in of V and the set Irrn(sl(2))ofisomorphism classes of n-dimensional representations of sl2. For example, the unique regular nilpotentconjugacy class corresponds to the unique n-dimensional irreducible representation of sl2.
From the results of section 5.4, we note that the e, h, f act on V via operators with trace zero. Hencethe above images of sl2 actually appear in sl(V ), the trace-zero subalgebra of End(V ). The bijectionsabove are a special case (for L = sl(V )) of the Jacobson-Morozov Theorem for semisimple Liealgebras L over F . See Bourbaki Groupes et Algebres de Lie book VIII, chapter 11.
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