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Math 55a Lecture Notes
This is Harvard College’s famous Math 55a, instructed by Dennis Gaitsgory. The formal name for this class is “Honors Abstract and Linear Algebra” but it generally goes by simply “Math 55a”. The permanent URL is http://web.evanchen.cc/~evanchen/coursework. html, along with all my other course notes.
- Fall Evan Chen
- 1 September 2, Contents
- 1.1 Boring stuff
- 1.2 Functions
- 1.3 Equivalence relations
- 2 September 4,
- 2.1 Review of equivalence relations go here
- 2.2 Universal property of a quotient
- 2.3 Groups
- 2.4 Homomorphisms
- 3 September 9,
- 3.1 Direct products
- 3.2 Commutative diagrams
- 3.3 Sub-things
- 3.4 Let’s play Guess the BS!
- 3.5 Kernels
- 3.6 Normality
- 3.7 Examples of normal groups
- 4 September 11,
- 4.1 Rings
- 4.2 Ring homomorphisms
- 4.3 Modules, and examples of modules
- 4.4 Abelian groups are Z-modules
- 4.5 Homomorphisms of R-modules
- 4.6 Matrices
- 4.7 Sub-modules and Ideals
- 5 September 16,
- 5.1 Review
- 5.2 Direct Sums of Modules
- 5.3 Direct Products of Modules
- 5.4 Sub-Modules Evan Chen (Fall 2014) Contents
- 5.5 Free Modules
- 5.6 Return to the Finite
- 6 September 18,
- 6.1 Linearly independent, basis, span
- 6.2 Dimensions and bases
- 6.3 Corollary Party
- 6.4 Proof of Theorem
- 7 September 23,
- 7.1 Midterm Solutions
- 7.2 Endomorphisms
- 7.3 Given a map we can split into invertible and nilpotent parts
- 7.4 Eigen-blah
- 7.5 Diagonalization
- 8 September 25,
- 8.1 Eigenspaces
- 8.2 Generalized eigenspaces
- 8.3 Spectral Theorem
- 8.4 Lemmata in building our proof
- 8.5 Proof of spectral theorem
- 8.6 Recap of Proof
- 9 September 30,
- 9.1 Review
- 9.2 Taking polynomials of an endomorphism
- 9.3 Minimal Polynomials
- 9.4 Spectral Projector
- 9.5 Polynomials
- 10 October 2,
- 10.1 Jordan Canonical Form
- 10.2 A big proposition
- 10.3 Young Diagrams
- 10.4 Proof of Existence
- 11 October 7,
- 11.1 Order of a Group
- 11.2 Groups of prime powers
- 11.3 Abelian groups and vector spaces are similar
- 11.4 Chinese Remainder Theorem
- 11.5 Not algebraically closed
- 12 October 9,
- 12.1 Group Actions
- 12.2 How do G-sets talk to each other?
- 12.3 Common group actions
- 12.4 More group actions
- 12.5 Transitive actions
- 12.6 Orbits Evan Chen (Fall 2014) Contents
- 12.7 Corollaries of Sylow’s Theorem
- 12.8 Proof of (b) of Sylow’s Theorem assuming (a)
- 13 October 14,
- 13.1 Proof of the first part of Sylow’s Theorem
- 13.2 Abelian group structure on set of modules
- 13.3 Dual Module
- 13.4 Double dual
- 13.5 Real and Complex Vector Spaces
- 13.6 Obvious Theorems
- 13.7 Inner form induces a map
- 14 October 16,
- 14.1 Artificial Construction
- 14.2 Orthogonal Subspace
- 14.3 Orthogonal Systems
- 14.4 Adjoint operators
- 14.5 Spectral theory returns
- 14.6 Things not mentioned in class that any sensible person should know
- 14.7 Useful definitions from the homework
- 15 October 21,
- 15.1 Generators
- 15.2 Basic Properties of Tensor Products
- 15.3 Computing tensor products
- 15.4 Complexification
- 16 October 23,
- 16.1 Tensor products gain module structure
- 16.2 Universal Property
- 16.3 Tensor products of vector spaces
- 16.4 More tensor stuff
- 16.5 Q & A
- 17 October 28,
- 17.1 Midterm Solutions
- 17.1.1 Problem
- 17.1.2 Problem
- 17.1.3 Problem
- 17.2 The space Λn sub(V )
- 17.3 The space Λn quot(V )
- 17.4 The Wedge Product
- 17.5 Constructing the Isomorphism
- 17.6 Why do we care?
- 18 October 30,
- 18.1 Review
- 18.2 Completing the proof that Λn sub(V ) = Λn quot(V )
- 18.3 Wedging Wedges
- 19 November 4, Evan Chen (Fall 2014) Contents
- 19.1 Representations
- 19.2 Group Actions, and Sub-Representations
- 19.3 Invariant Subspaces
- 19.4 Covariant subspace
- 19.5 Quotient spaces and their representations
- 19.6 Tensor product of representations
- 20 November 6,
- 20.1 Representations become modules
- 20.2 Subrepresentations
- 20.3 Schur’s Lemma
- 20.4 Splittings
- 20.5 Table of Representations
- 20.6 Induced and Restricted Representations
- 21 November 11,
- 21.1 Review
- 21.2 Homework Solutions
- 21.3 A Theorem on Characters
- 21.4 The Sum of the Characters
- 21.5 Re-Writing the Sum
- 21.6 Some things we were asked to read about
- 22 November 13,
- 22.1 Irreducibles
- 22.2 Products of irreducibles
- 22.3 Regular representation decomposes
- 22.4 Function invariants
- 22.5 A Concrete Example
- 23 November 18,
- 23.1 Review
- 23.2 The symmetric group on five elements
- 23.3 Representations of S 5 /(S 3 × S 2 ) – finding the irreducible
- 23.4 Secret of the Young Diagrams
- 23.5 The General Theorem
- 24 November 20,
- 24.1 Reducing to some Theorem with Hom’s
- 24.2 Reducing to a Combinatorial Theorem
- 24.3 Doing Combinatorics
- 25 December 2,
- 26 December 4,
- 26.1 Categories
- 26.2 Functors
- 26.3 Natural Transformations
§1 September 2, 2014
§1.1 Boring stuff
Sets include R, Z, et cetera. A subset Y ⊆ X is exactly what you think it is. Q, { 0 }, { 1 }, ∅, R ⊆ R. Yay. X 1 ∪ X 2 , X 1 ∩ X 2.
... Gaitsgory what are you doing For a fixed universe X, we write Y , X \ Y , X − Y for {x ∈ X | x /∈ Y }.
Lemma 1. For Y ⊂ X, (^) ( Y
= Y.
Proof. Trivial. darn this is being written out?
x ∈
Y
⇐⇒ x /∈ Y ⇐⇒ x ∈ Y.
Hence
Y
= Y.
Lemma 1. (X 1 ∩ X 2 ) = X 1 ∪ X 2.
Proof. Compute
x ∈ X 1 ∩ X 2 ⇐⇒ x /∈ X 1 ∩X 2 ⇐⇒ x /∈ X 1 ∨x /∈ X 2 ⇐⇒ x ∈ X 1 ∨x ∈ X 2 ⇐⇒ x ∈ X 1 ∪X 2.
Lemma 1. X 1 ∪ X 2 = X 1 ∩ X 2.
Proof. HW. But this is trivial and follows either from calculation or from applying the previous two lemmas.
Given a set X we can consider its power set P(X). It has 2n^ elements.
§1.2 Functions
Given two sets X and Y a map (or function) X f −→ Y is an assignment ∀x ∈ X to an element f x ∈ Y. Examples: X = {55 students}, Y = Z. Then f (x) = $ in cents (which can be negative).
Definition 1.4. A function f is injective (or a monomorphism) if x 6 = y =⇒ f x 6 = f y.
Definition 1.5. A function f is surjective (or an epimorphism) if ∀y ∈ Y ∃x ∈ X : f x = y.
Composition; X f −→ Y g −→ Z.
§2 September 4, 2014
§2.1 Review of equivalence relations go here
meow
§2.2 Universal property of a quotient
X
f (^) - Y
X/ ∼
π ?
˜f
Proposition 2. Let X and Y be sets and ∼ an equivalence relation on X. Let f : X → Y be a function which preserves ∼, and let π denote the projection X → X/ ∼. Prove that there exists a unique function f˜ such that f = f˜ ◦ π.
The uniqueness follows from the following obvious lemma.
X
f (^) - Y
X′
g ?
f^2
f^1
Lemma 2. In the above commutative diagram, if g is surjective then f 1 = f 2.
Proof. Just use g to get everything equal. Yay.
§2.3 Groups
Definition 2.3. A semi-group is a set G endowed with an associative^1 binary operation ∗ : G^2 → G.
Lots of groups work.
Example 2.4 (Starfish) Let G be an arbitrary set and fix a g 0 ∈ G. Then let ab = g 0 for any a, b ∈ G. This is a semigroup.
A “baby starfish” has |G| = 1.
(^1) (ab)c = a(bc)
Definition 2.5. A semi-group G is a monoid if there exists an identity 1 ∈ G such that ∀g ∈ G, g · 1 = 1 · g = g.
Proposition 2. The identity of any semi-group G is unique.
Proof. Let 1, 1′^ be identities. Then
1 = 1 · 1 ′^ = 1′.
Definition 2.7. A group is a monoid G with inverses: for any g ∈ G there exists g−^1 with gg−^1 = g−^1 g = 1.
Proposition 2. Inverses are unique.
Proof. Suppose x 1 , x 2 are both inverses of g. Then
x 1 = x 1 gx 2 = x 2.
Definition 2.9. A group is abelian if it is commutative.
§2.4 Homomorphisms
Definition 2.10. Let G and H be groups. A group homomorphism is a map f : G → H that preserves multiplication.
§3.4 Let’s play Guess the BS!
“In what follows I’ll state some false statements. You will be asked to prove them at your own risk.”
Lemma 3. Given ϕ : G → H a homomorphism, ϕ(G) is a subgroup of H.
Proof. Given any ϕ(a), ϕ(b) in ϕ(G) we have ϕ(a)ϕ(b) = ϕ(ab) ∈ ϕ(G). Then use ϕ(1) = 1 to get the rest of the conditions.
Lemma 3. If ϕ : G → H and H′^ ≤ H, then ϕ−^1 (H′) is a subgroup of G.
Proof. This one turns out to be true.
Fact 3.6. Given H 1 , H 2 subgroups of G, H 1 ∪ H 2 need not be a subgroup of G. Proof. Take G = Z, H 1 = 100Z, H 2 = 101Z.
§3.5 Kernels
Definition 3.7. Given a homomorphism ϕ : G → H, the kernel ker ϕ is defined by ker ϕ = ϕ−^1 ({ 1 }).
Proposition 3. Let ϕ : G → H be a homomorphism. Then ϕ is injective as a map of sets if and only if ker ϕ = { 1 }.
Proof. In all cases 1 ∈ ker ϕ. If |ker ϕ| 6 = 1 then clearly ϕ is not injective. On the other hand, suppose ker ϕ = { 1 }. If ϕa = ϕb we get ϕ(ab−^1 ) = 1, so if we must have ab−^1 = 1 or a = b.
Definition 3.9. Let G be a group and let H ≤ G be a subgroup. We define the right equivalence relation on G with respect to H ∼r^ as follows: g 1 ∼r^ g 2 if ∃h ∈ H such that g 2 = g 1 h. Define ∼`^ similarly. To check this is actually an equivalence relation, note that 1 ∈ H =⇒ g ∼r^ g and g 1 ∼ g 2 =⇒ g 1 = g 2 h =⇒ g 1 h−^1 = g 2 =⇒ g 2 ∼r^ g 1. Finally, if g 1 = g 2 h′^ and g 2 = g 3 h′′^ then g 1 = g 3 (h′h′′), so transitivity works as well. Note that g 1 ∼r^ g 2 ⇐⇒ g 1 − 1 g 2 ∈ H. Definition 3.10. Let G/H be the set of equivalence classes of G with respect to ∼r.
§3.6 Normality
Definition 3.11. Let H be a subgroup of G. We say H is normal in G if ∀g ∈ G and ∀h ∈ H we have ghg−^1 ∈ H. (This called the conjugation of h by g.)
Theorem 3. Let H be a normal subgroup of G. Consider the canonical projection π : G → G/H. Then we can place a unique group structure on G/H for which π is a homomorphism.
Proof. For uniqueness, apply the lifting lemma to the following commutative diagram
G × G
G
π (^) - G/H
G/H × G/H
π × π ? ...............................................
f^
Now we claim existence when H is normal. Let’s ignore normality for now. Given x, y ∈ G/H, we choose x′y′^ = π(xy) for x ∈ π−^1 (x′) and y ∈ π−^1 (y′). Now to check this is well-defined (and makes the diagram commutative). Given π(x 1 ) = π(x 2 ) = x′^ and π(y 1 ) = π(y 2 ) = y′, we want to check that
π(x 1 · y 1 ) = π(x 2 · y 2 ).
Evidently x 2 = x 1 h′^ and y 2 = y 1 h′′, so the above would be equivalent to
π(x 1 · y 1 ) = π(x 1 h′y 1 h′′).
You can check this is equivalent to
y 1 − 1 h′y 1 ∈ H
for all choices of y 1 ∈ G and h′^ ∈ H. That’s where (and only where) the normal condition comes in. It implies that our map is indeed well-defined, and we win. Finally, we need to show associativity and inverses. We want the following diagram to commute.
G^3
G/H × G/H × G/H
id × ∗-
π (^3)
G/H × G/H
∗ × id ? ∗
- G/H
?
Note that the G^3 has been added in. We use associativity of G to do cool things. OK the rest of the details are left as an exercise.
§4 September 11, 2014
Happy birthday to Max Schindler!
§4.1 Rings
Definition 4.1. A ring R is a set endowed with two binary operations + and ·, addition and multiplication, such that
(i) R is an abelian group with respect to addition. The additive identity is 0. (ii) R is a monoid^2 with respect to multiplication, whose identity is denoted 1.
(iii) Multiplication distributes over addition.
The ring R is commutative if multiplication is commutative as well.
Example 4. Here are examples of commutative rings:
- Z, R are rings.
- R[t 1 , t 2 ,... , tn] are rings.
- The integers modulo n are rings.
Example 4. Square matrices are the standard example of non-commutative rings.
Lemma 4. Let R be a ring. Then r · 0 = 0.
Proof. Compute r · 0 = r · (0 + 0) = r · 0 + r · 0.
Hence r · 0 = 0.
Lemma 4. In a ring, r · (−1) = −r.
Proof. Compute
r · (−1) + r = r · (−1) + r · 1 = r · (−1 + 1) = r · 0 = 0.
A little sidenote made at the end of class.
Definition 4.6. A commutative ring R is a field if R − { 0 } is a group with respect to multiplication. (^2) Some sources do not require a 1 to exist, but in practice no one cares about rings without 1 anyway.
§4.2 Ring homomorphisms
As usual, we figure out how two rings talk to each other.
Definition 4.7. Given rings R and S, a ring homomorphism is a function ϕ : R → S which respects addition and multiplication such that ϕ(1) = 1.
Example 4. We can embed Z in R, Q in R, and so on. These maps will all be homomorphisms. Moreover, we can compose ring homomorphisms.
Example 4. We can construct a homomorphisms from R[t] into R by sending p(t) to p(2014).
§4.3 Modules, and examples of modules
In this section, fix a ring R. The addition is + and has identity 0; the multiplication has identity 1 and is written r 1 r 2.
Definition 4.10. A left R-module is an additive abelian group M (meaning M is an abelian group with operation +) equipped with an additional multiplication: for each r ∈ R and m ∈ M we define an r · m ∈ M. This multiplication must satisfy the following properties for every r 1 , r 2 ∈ R and m ∈ M :
(i) r 1 · (r 2 m) = (r 1 r 2 ) · m.^3
(ii) Multiplication is distributive, meaning (r 1 +r 2 )·m = r 1 ·m+r 2 ·m. and r·(m 1 +m 2 ) = r · m 1 + r · m 2.
(iii) 1 · m = m.
A module generalizes the idea of vector spaces.
Example 4. A trivial example of a module is M = { 0 }. All the axioms of modules are identities inside M , so there is nothing to verify.
Example 4. Additionally, we can let M be additive abelian group underlying R. The action of M on R is just left multiplication.
(^3) Note that if R is not commutative, then it is not necessarily possible to define a right R-module by simply setting m · r = r · m. In other words, left and right modules are different beasts.
Proposition 4. The above map from HomR (Rm, Rn) to Matn×m(R) is a bijection of sets.
Here HomR (Rm, Rn) is the set of ring homomorphisms. In other words, with T a homomorphism, our T is determined uniquely by T (e 1 ), T (e 2 ),... , T (en).
Proof. First, suppose M is given. We will construct T from M. Obviously we will need to have
T (ei) =
mi 1 mi 2 .. . min
in other words, the ith column of M. If T is going to be a homomorphism, we had better have
T (〈r 1 , r 2 ,... , rm〉) = T (r 1 e 1 + r 2 e 2 + · · · + rmem) =
∑^ m
i=
riT (ei).
Hence we know exactly what T needs to be based on M. Hence, we just have to show it is actually a homomorphism, which is not hard. Conversely, it is trivial to produce M given T.
§4.7 Sub-modules and Ideals
Let M be an R-module.
Definition 4.18. A left R-submodule of M is a subset M ′^ ⊆ M such that
- M ′^ is a subgroup (with respect to inherited addition).
- If m′^ ∈ M ′^ then ∀r ∈ R, we have rm′^ ∈ M ′. In other words, M ′^ absorbs left multiplication.
Definition 4.19. A left ideal in I is a subset of R which is also a left R-submodule under the natural interpretation. Explicitly, I is a left ideal if and only if
(i) I is closed under addition, meaning i 1 + i 2 ∈ I for all i 1 , i 2 ∈ I,
(ii) 0 ∈ I, and i ∈ I =⇒ −i ∈ I,
(iii) For any r ∈ R and i ∈ I, ri ∈ I.
A right ideal is defined similarly, where the last axiom has ir ∈ I (as opposed to ri ∈ I). If I is both a left and right ideal, it is a two-sided ideal or just ideal.
Example 4. If R = Z, then { 0 }, 2Z, 999Z, and Z are all submodules.
§5 September 16, 2015
§5.1 Review
Below is problem 10 on the previous problem set, which is apparently important.
Proposition 5. For any R-module M , the map from HomR(R, M ) to M by α 7 → α(1) is an isomor- phism. In other words, a homomorphism of R to an R-module is determined by its value at 1.
Also we have the following obvious bijections. We have a map
Hom(X 1 ∪ X 2 , Y ) → Hom(X 1 , Y ) × Hom(X,Y )
by α 7 → (α ◦ i 1 , α ◦ i 2 ). X 1 t X 2
X 1
i^1
X 2
i 2
We have another map
Hom(X, Y 1 × Y 2 ) → Hom(X, Y 1 ) × Hom(X, Y 2 )
by β 7 → (p 1 ◦ β, p 2 ◦ β). Y 1 × Y 2
Y 1
π^1
Y 2
π 2
§5.2 Direct Sums of Modules
The analog of these two maps in the world of modules is the direct set.
Definition 5.2. Let M 1 and M 2 be R-modules. We define the direct sum, denoted M 1 ⊕ M 2 , as the module whose underlying group is M 1 × M 2. The action of R on M 1 ⊕ M 2 is r · (m 1 , m 2 ) = (rm 1 , rm 2 ).
Definition 5.3. If M is an R-module, then M ⊕n^ represents the direct product of M n times.
Proposition 5. There is a bijection
HomR
N,
a
Ma
a
Hom(N, Ma)
by composition with pa.
However, surprisingly enough, it is NOT true that we have a bijection
HomR
a
Ma, N
a
HomR(Ma, N ).
We have to make the following modification.
Definition 5.7. We define the infinite direct sum
a Ma, a subset of^
a Ma, in which at most finitely many indices a have a nonzero value.
Proposition 5. There is a bijection
HomR
a
Ma, N
a
HomR(Ma, N ).
by pre-composition with inclusion.
Note that
a Ma^ =^
a Ma^ when we have finitely many indices.
§5.4 Sub-Modules
Let M ′^ be a submodule of a module M. Then we have a projection π : M → M/M ′.
Lemma 5. The set M/M ′^ acquires a unique structure of an R-module so that π is a homomor- phism of R-modules.
Proof. We already know we can put the necessary group structure. So we first have to show there is a map act′ r such that the following diagram commutes.
M
π- M/M ′
M
actr ? π
- M/M ′
act′ r ?
π (^) ◦ (^) act r
The universal property (Proposition 2.1) implies that act′ r exists and is unique if and only if π ◦ actr vanishes on M ′. Note that
π ◦ actr(m′) = π(r · m′) = 0
since m′^ ∈ M ′^ =⇒ r · m′^ ∈ M ′, end of story. The axioms on act′ r follow by inheritance from M because we commutative diagram are magic. For example, if you want to show associativity, that’s equivalent to
actr 1 r 2 = actr 2 ◦ actr 1.
If I plug these two things into the dotted arrow, both make the diagram commute. That’s enough to make them equal. (This was Lemma 2.2.)
Lemma 5. Let M ′^ be a R-submodule of the R-module M. Let T : M → N be an R-module homomorphism such that T (M ′) = 0. Then there is a unique T ′^ : M/M ′^ → N such that T ′^ ◦ π = T.
M
T -
N
M/M ′
π ?
T^
′^
Proof. We already know T ′^ is a homomorphism of groups. So M has an actR, while M/M ′^ also has an action act′ R. We wish to show that
act′ R ◦T ′^ = T ◦ actR.
Consider the following commutative diagram.
M/M ′
M/M ′^
π
act
′ R
M
T -
N
T ′
N
act
R
T ′
Do magic. Again, verifying axioms is just a matter of the “both satisfy the dotted arrow” trick in the previous proof.
§5.5 Free Modules
Definition 5.11. Let R⊕A^ be the free module, which is the set of maps { A ϕ −→ R | ϕ(a) = 0 except for finitely many a
