Download Gordon Brown Spring 2016 Background Notes for Quiver

Gordon Brown
Spring 2016
Background Notes for Quiver Representations
1. Fields
Since every vector space is “over a field,” it’s good to know a little bit about fields themselves.
Roughly speaking, a field is a collection of objects on which there are two operations, which have the
same properties as addition and multiplication of real numbers, respectively. More specifically:
Definition: A field is a set K1 with two operations, called addition (denoted +) and multiplication (denoted by
concatenation without space), subject to the following axioms.
• Closure: For every α, β ∈ K, the sum α + β and the product αβ are also elements of K.
• Associativity: (α + β) + γ = α + (β + γ) and α(βγ) = (αβ)γ for all α, β, γ ∈ K.
• Commutativity: α + β = β + α and αβ = βα for all α, β ∈ K.
• Additive Identity: There exists an element 0 ∈ K such that α + 0 = α for all α ∈ K.
• Multiplicative Identity: There exists an element 1 ∈ K such that α 1 = α for all α ∈ K.
• Additive Inverses: For every α ∈ K, there exists an element −α ∈ K such that α + (−α) = 0.
• Multiplicative Inverses: For every α ∈ K, α , 0, there exists an element α−1 ∈ K such that αα−1 = 1.
• Distributivity: For every α, β, γ ∈ K, we have α(β + γ) = αβ + αγ.
Note that additive and multiplicative inverses are unique (why?).
Examples: The real numbers R. The rational numbers,
Q=
m
: m, n are integers
n
and the complex numbers
C = {x + iy : x, y ∈ R}
where i =
√
−1 is the imaginary unit are also fields. Since Q ⊂ R ⊂ C, Q is a subfield of R and R is a subfield
of C. By transitivity, Q is in turn a subfield of C.
Of course, Q , R and R , C. An example of an element in R but not in Q is
√
2. An example of an
element in C but not in R is, of course, i.
The integers Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . . }, however, do not make a field (why?). They’re something
else, a ring.
1 The
notion of a field was formalized mostly by German mathematicians, who used the German word for body – Körper – to mean
field. This explains why the letter K is often used for fields instead of F.
1
It’s amazing how many mathematical documents begin with the sentence “let K be an algebraically
closed field of characteristic zero.” Here’s some background on algebraic closure and characteristic.
Definition: A field K is algebraically closed if every non-constant polynomial p(x) with coefficients in K has a
root in K. In symbols, K is algebraically closed if for every p(x) = αn xn + αn−1 xn−1 + · · · + α1 x + α0 with αi ∈ K
for all i and n ≥ 1, there exists an element β ∈ K such that p(β) = 0. (It is in fact true that such a p(x) with
αn , 0 will have exactly n roots in K counting multiplicities if K is algebraically closed.)
Examples: Neither Q nor R is algebraically closed (why?), but one of the most celebrated results of classical
algebra, often called the fundamental theorem of algebra, states that C is algebraically closed. As you may
have guessed, it is in fact true that every field is contained in an algebraically closed field.
Definition: The characteristic of a field K is the smallest positive integer n such that the n-fold sum
1 + 1 + 1 + · · · + 1 is equal to 0. This is denoted char K = n. If no such positive integer n exists, the
characteristic is defined to be zero.
Examples: The fields Q, R, and C all have characteristic zero. Note that if a field’s characteristic is n > 0,
then n must be a prime number (why?).
2. Vector Spaces
In physics, we define a vector to be a quantity which has both a magnitude and a direction. In mathematics,
however, the notions of magnitude and direction are often discarded in favor of generalizing to the more
abstract notion of quantities which can be added together and scaled by the elements in a field.
Definition: Let K be a field. A vector space over K, or a K-vector space, is a set V with two operations, called
vector addition (denoted +) and scalar multiplication (denoted by concatentation without space), subject to the
following axioms.
• Closure: For every v, w ∈ V and α ∈ K, the sum v + w and the scalar multiple αv are also elements of V.
• Associativity: (u + v) + w = u + (v + w) for all u, v, w ∈ V.
• Commutativity: v + w = w + v for all v, w ∈ V.
• Zero Vector: There exists an element 0 ∈ V such that 0 + v = v for all v ∈ V.
• Scalar Identity: 1 v = v for all v ∈ V.
• Additive Inverses: For every v ∈ V, there exists an element −v ∈ V such that v + (−v) = 0.
• Distributivity: α(v + w) = αv + αw and (α + β)v = αv + βv for all v, w ∈ V and α, β ∈ K.
• Compatibility: α(βv) = (αβ)v for all v ∈ V and α, β ∈ K.
2
In practice, it’s helpful to think of a vector space as consisting of all possible linear combinations of the
vectors in some basis. Here’s a more exotic vector space that I’ve encountered in my research.
Example: For a positive integer n, a partition of n is a finite sequence λ = (λ1 , . . . , λm ) of positive integers
λ1 , . . . , λm whose sum is n, λ1 + · · · + λm = n. The phrase “λ partitions n” is denoted symbolically by λ ` n.
By convention, partitions are always in a non-increasing order, λ1 ≥ · · · ≥ λm . So for example, the partitions
of 4 are (4), (3, 1), (2, 2), (2, 1, 1), and (1, 1, 1, 1).
Partitions can be represented pictorially by Young frames.2 The Young frame corresponding to a partition
λ = (λ1 , . . . , λn ) ` m is a left-justified stack of m empty squares. There are n rows of squares, with λi squares
in the i th row. So for example, the Young frames corresponding to the partitions of 4 listed above are
,
,
,
,
and
.
For every n, define Yn to be the C-vector space spanned by the Young frames of size n. In other words,
the Young frames of size n are vectors in, and form a basis of, Yn . So for example, vectors in Y4 look like
α1
+ α2
+ α3
+ α4
+ α5
where α1 , . . . , α5 ∈ C. Hence the dimension of Yn is the number of partitions of n, e.g., dim Y4 = 5.
3. Linear Maps
Mathematics is rife with many different types of objects, each endowed with different types of structure.
Always of interest are the functions which go between objects of the same type and preserve the structure.
A linear transformation, or linear map, is nothing more than a function between K-vector spaces which
preserves their vector space structure. Since vector spaces have the structure of a set endowed with the
operations of vector addition and scalar multiplication, a linear map should consist of a well-defined set
function which is somehow “compatible” with adding and scaling. The following definition formalizes this.
Definition: Let K be a field, and V and W K-vector spaces. A linear map is a function f : V → W such that
• f (v1 + v2 ) = f (v1 ) + f (v2 ) for all v1 , v2 ∈ V and
• f (αv) = α f (v) for all α ∈ K and v ∈ V.
The latter two conditions are collectively referred to as linearity.
An important fact about linear maps f : V → W is that they’re defined on all of V if and only if they’re
defined on just a basis of V. Indeed, suppose {v1 , . . . , vn } is a basis for V and we’ve already defined what
the vectors f (v1 ), . . . , f (vn ) are in W. Then given any other v ∈ V, since v can be expressed (uniquely!) as
2 Named
for Alfred Young (1873 - 1940), an English mathematician.
3
v = α1 v1 + · · · + αn vn for some α1 , . . . , αn ∈ K, and we declared f to be a linear map, there is no choice for
what f (v) can be except
f (v) = f (α1 v1 + · · · + αn vn ) = f (α1 v1 ) + · · · + f (αn vn ) = α1 f (v1 ) + · · · + αn f (vn ).
So defining a linear map f : V → W is as easy as 1, 2, 3:
1. Pick a basis {v1 , . . . , vn } of V.
2. Choose vectors f (v1 ), . . . , f (vn ) ∈ W.
3. Declare that f is obtained from these assignments by “linearly extending,” or “extending by linearity.”
Example: Let Yn be as previously. Define a linear map addn : Yn → Yn+1 by linearly extending the following
definition on the basis vectors of Yn , that is, on the Young frames λ of size n.
X
addn (λ) =
µ
µ=λ+
For example, if λ =
in Y3 , then
=
add3
+
+
since the Young frames µ in Y4 which can be obtained from
by adding a single box are
,
, and
.
Similarly, define a linear map rem : Yn → Yn−1 (this only works if n ≥ 2) by linearly extending the definition
X
remn (λ) =
ν
ν=λ−
so that, for example,
rem3
=
+
.
Notice that the composite maps addn ◦ remn+1 and remn ◦ addn−1 go from Yn to itself.
Exercise: Are either of the composite maps addn ◦ remn+1 and remn ◦ addn−1 equal to the identity map on Yn ?
Below is an illustration that might help.
4
It’s called the Young lattice, a graph whose vertices are Young frames (we could assign each edge a
direction and make it a quiver, but let’s not right now). Think of the Young lattice as having one row for
each size n of Young frame, so the Young frames in row n form a basis of the vector space Yn . This picture
only shows the first 5 rows of the Young lattice (including size zero); the pattern continues upwards forever.
Exercise:
• Under what condition will two Young frames have an edge connecting them in the Young lattice?
• Reformulate the definitions of the linear maps addn and remn in terms of edges in the Young lattice.
A quick detour before proceeding: linear maps f : V → W can be scaled and added to other linear maps
g : V → W in natural ways.
Definition: Let Hom(V, W) denote the collection of all linear maps V → W.3
Exercise:
• Show that Hom(V, W) is a vector space, or at least, that it has well-defined scalar multiplication and
vector (linear map) addition operations.
• The linear map (remn ◦ addn−1 ) − (addn ◦ remn+1 ) is equal to a familiar linear map Yn → Yn . What is it?
Note that Hom(V, V) also has the operation of composition of linear maps, giving it the structure of
something more than just a vector space, an algebra. If time permits we’ll return to this phenomenon.
4. Direct Sums
Let V be a vector space, and W1 and W2 subspaces of V (so they’re vector spaces in their own right but
contained in V). We can always form the sum
W1 + W2 = {w1 + w2 : w1 ∈ W1 , w2 ∈ W2 },
making another subspace of V. Two natural questions arise here.
• Is V = W1 + W2 ? That is, we already know V ⊇ W1 + W2 , but is V ⊆ W1 + W2 ?
• Does the sum W1 + W2 have any redundancy? That is, are there any nonzero vectors in W1 ∩ W2 ?
Answering the first question amounts to answering whether or not every v ∈ V can be expressed v = w1 + w2
for some w1 ∈ W1 and w2 ∈ W2 . To relate this to the second question, let’s think in terms of bases.
Suppose {w1 , . . . , wm } is a basis of W1 and {w01 , . . . , w0n } is a basis of W2 . We might ask whether or not the
union of the two, {w1 , . . . , wm , w01 , . . . , w0n }, is a basis for all of V. Recall that for {w1 , . . . , wm , w01 , . . . , w0n } to be
a basis of V means
3 “Hom”
is an abbreviation for “homomorphism.”
5
• the span of its elements is all of V and
• its elements are linearly independent.
But saying the span of {w1 , . . . , wm , w01 , . . . , w0n } is all of V is equivalent to saying the answer to the first
question is yes. Meanwhile saying w1 , . . . , wm , w01 , . . . , w0n are linearly independent is equivalent to saying
the answer to the 2nd question is no, W1 ∩ W2 = {0} (why?). Let’s summarize these observations.
Proposition: Let V be a vector space with subspaces W1 and W2 . Suppose {w1 , . . . , wm } is a basis of W1 and
{w01 , . . . , w0n } is a basis of W2 .
1. V = W1 + W2 if and only if {w1 , . . . , wm , w01 , . . . , w0n } spans V.
2. W1 ∩ W2 = {0} if and only if w1 , . . . , wm , w01 , . . . , w0n are linearly independent.
3. {w1 , . . . , wm , w01 , . . . , w0n } is a basis of V if and only if both statements 1 and 2 are true.
Definition: The sum W1 + W2 is a direct sum if either condition of statement 2 in the proposition is satisfied.
That a sum is direct may be indicated by writing it as W1 ⊕ W2 .
Note that if V = W1 ⊕ W2 , then dim V = dim W1 + dim W2 by statement 3 of the proposition, but the
converse is not true. For a counterexample, set V = R2 and let W1 = W2 be the subspace spanned by h1, 1i.
Then dim R2 = 2 = 1 + 1 = dim W1 + dim W2 , but R2 , W1 + W2 and in particular the sum isn’t even direct.
Example: Let Y be the vector space spanned by the set of all Young frames, of all sizes. So a basis for Y is
∞
M
the set of all the Young frames in the Young lattice (so Y is infinite-dimensional). Then clearly Y =
Yn .
n=1
Just as there are direct sums of vector spaces, there are direct sums of linear maps.
Definition: If f1 : V1 → W1 and f2 : V2 → W2 are linear maps, and the sums V1 ⊕ V2 and W1 ⊕ W2 are direct,
then we have a linear map f1 ⊕ f2 : V1 ⊕V2 → W1 ⊕W2 given by linearly extending f1 ⊕ f2 (v1 +v2 ) = f1 (v1 )+ f2 (v2 ).
Example: Define the linear maps add : Y → Y and rem : Y → Y by
add =
∞
M
addn
and
rem =
∞
M
remn .
n=2
n=1
So add just applies addn to the nth row of the Young lattice for every n simultaneously, while rem does the
same with remn .
6
5. Eigenvalues
From now on, we view an n × n matrix A with coefficients in K as a linear map A : Kn → Kn in the usual
way: A sends a column vector v ∈ Kn to the column vector obtained by doing the matrix multiplication Av.
Recall that an eigenvector of A is a vector v ∈ Kn which is just scaled by A, so Av = λv for some λ ∈ K.4
The scalar λ is the eigenvalue corresponding to the eigenvector v. Recall also that the eigenvalues of A are
the zeros of the characteristic polynomial det(A − λIn ) where

1





In = 





1
..
.
1
1







 = diag(1, 1, . . . , 1, 1)





is the n × n identity matrix.5 In practice, we’ll work with matrices A over the field R of real numbers, so
the entries of A will be in R and A : Rn → Rn as a linear map. But since R is not algebraically closed its
eigenvalues may be complex numbers, so if we’re being careful A is really a linear map Cn → Cn .
Exercise: Show that eigenvectors of different eigenvalues are linearly independent.
For an eigenvalue λ of the matrix A, let Vλ be the subspace of Cn spanned by all the eigenvectors with
eigenvalue λ, the λ-eigenspace of A. Since eigenvalues are zeros of a degree n polynomial, there can be at
most n of them (although there are exactly n of them if you count multiplicities). By the exercise, if λ1 , . . . , λm
are the eigenvalues of A (so m ≤ n), then the sum Vλ1 + · · · + Vλm is direct and we have
Cn ⊇ Vλ1 ⊕ · · · ⊕ Vλm .
A natural question to ask is whether Cn ⊆ Vλ1 ⊕ · · · ⊕ Vλm , that is, whether or not Cn has a basis consisting
of eigenvectors of A.
First of all, if A is a diagonal matrix then Cn has a basis of eigenvectors of A. For example, suppose

1



A = 








 = diag(1, 2, 2, 3).



2
2
3
Let {e1 , e2 , e3 , e4 } be the standard basis of C4 , that is,
 
1
 
0
e1 =   ,
0
 
 
0
4 The
 
0
 
1
e2 =   ,
0
 
 
0
 
0
 
0
e3 =   ,
1
 
 
0
 
0
 
0
e4 =   .
0
 
 
1
prefix “eigen” comes from German; roughly speaking it means “own” or “self.”
left blank in a matrix are always taken to be zero.
5 Entries
7
Then multiplying it out or just looking at the columns of A, we see that
Ae1 = e1 ,
Ae2 = 2e2 ,
Ae3 = 2e3 ,
Ae4 = 3e4 ,
so C4 has a basis of eigenvectors of A. (Also, the determinant of a diagonal matrix is the product of its
diagonal entries, so the zeros of the characteristic polynomial det(A − λIn ) are just the diagonal entries of A.)
Notice that an eigenvalue can occur multiple times; this is precisely its multiplicity as a zero of det(A − λIn )
and in turn the dimension of the corresponding eigenspace.
Exercise: Show that if the standard basis vectors e1 , . . . , en of Cn are all eigenvectors of A, then A is diagonal.
So we have the following.
Proposition: An n×n matrix A is diagonal if and only if the standard basis vectors e1 , . . . , en of Cn are eigenvectors of A.
Returning to the earlier question, we see that A being diagonal means not only does Cn have a basis of
eigenvectors of A, but it’s the standard basis in particular. But there are lots of other bases for Cn than just
the standard one, so how can we tell if the eigenvectors of A make some other basis of Cn ?
Exercise:
• Let f : V → W be a linear map, and {v1 , . . . , vn } a basis for V. Show that { f (v1 ), . . . , f (vn )} is a basis of
W if and only if f is invertible, that is, if there exists a linear map f −1 : W → V with f ◦ f −1 = IdW and
f −1 ◦ f = IdV . (Equivalently, f is an isomorphism, that is, a linear map which is bijective.)
• Conclude that if M is an n × n matrix, then the columns of M are linearly independent if and only if M
is invertible, that is, if there exists an n × n matrix M−1 with M M−1 = M−1 M = In .
Suppose now that {v1 , . . . , vn } are eigenvectors of A which form a basis of Cn , so Avi = λi vi for some
λi ∈ C. (The λi might have repeats, but the vi are distinct.) Construct an n × n matrix M by making its i th
column the vector vi . In other words, as a linear map M sends the standard basis vector ei to vi for each i.
By the exercise, M is invertible.
Exercise: Show that e1 , . . . , en are eigenvectors of the matrix MAM−1 . Conclude that MAM−1 is diagonal.
So if Cn has a basis of eigenvectors of A, then there’s an invertible matrix M such that M−1 AM is diagonal.
On the other hand, if there’s a square matrix N such that N−1 AN is diagonal, say N−1 AN = D, we’d
like to know if Cn has a basis of eigenvectors of A. Since D is diagonal, e1 , . . . , en are eigenvectors of D, say
Dei = λi ei for some λi ∈ C. But since N−1 AN = D,
N−1 A N ei
=
λi ei
N N−1 A N ei
=
N(λi ei )
A(N ei ) =
λi (N ei )
8
and the last line shows that Nei is an eigenvector of A with eigenvalue λi for each i. In fact, {Ne1 , . . . , Nen }
forms a basis of Cn (why?), so Cn does have a basis of eigenvectors of A. Eureka! Let’s summarize.
Theorem: Let A be an n × n matrix with entries in C. Then Cn has a basis of eigenvectors of A if and only if there
exists an invertible matrix M such that M−1 AM is diagonal.
Definition: A square matrix A is diagonalizable if either condition of the theorem holds.
Definition: Two square matrices A and B are similar, or conjugate, if there exists an invertible matrix N such
that A = N−1 BN. (So diagonalizable means similar to a diagonal matrix.)
6. The Jordan Form
An n × n matrix A has n linearly independent eigenvectors precisely when it’s diagonalizable. What if A
isn’t diagonalizable, that is, it doesn’t have n linearly independent eigenvectors? Is there a next-best-thing?
Fortunately the answer is yes. For starters, the next-best-thing to an eigenvector is a generalized eigenvector.
Definition: Let A be an n × n matrix with entries in C. A generalized eigenvector of A with eigenvalue λ ∈ C
is a vector v ∈ Cn such that (A − λIn )r v = 0 for some positive integer r.
Note that bona fide eigenvectors are also generalized eigenvectors: they just need r = 1 here.
Example:

3
• Show that the only eigenvalue of B = 
0

1
 is λ = 3, but e1 is its only eigenvector up to scaling.
3
• Though B isn’t diagonalizable, show that e2 is a generalized eigenvector with eigenvalue λ = 3.
Example:

2


• Show that the only eigenvalue of C = 0

0
1
2
0

0

1 is λ = 2, but e1 is its only eigenvector up to scaling.

2
• Though C isn’t diagonalizable, show that e2 and e3 are generalized eigenvectors with eigenvalue λ = 2.
Definition: For λ ∈ C, let Jλ,n be the n × n matrix

λ





Jλ,n = 





1
λ
1
..
.
..
.
λ
9







 ,


1 

λ
the Jordan block with eigenvalue λ of size n. (In the exercises, B = J3,2 and C = J2,3 .)
The exercises hint that for every Jordan block Jλ,n , the standard basis vectors e1 , . . . , en of Cn are all
generalized eigenvectors of Jλ,n with eigenvalue λ. But a square matrix can have many different eigenvalues,
so based on past experience it’s natural to define the following.
Definition: A square matrix J is in Jordan form, or is a Jordan matrix, if it has Jordan blocks Jλi ,mi with different
eigenvalues λi and sizes mi along the diagonal and zeros elsewhere:


 Jλ1 ,m1





Jλ2 ,m2


 = diag( Jλ1 ,m1 , Jλ2 ,m2 , . . . , Jλr ,mr ).
J = 
.

..






Jλr ,mr
For example, the matrix

3







A = 







1
3
−4
−4
2
1
2









 = diag( J3,2 , J−4,1 , J−4,1 , J2,3 )




1

2
is in Jordan form. The exercises further hint that for every Jordan matrix J, the standard basis vectors
e1 , . . . , en of Cn are all generalized eigenvectors of J. But again, there are lots of other bases for Cn , so our
past experience tells us to now pose the question: is every square matrix A similar to a Jordan matrix?
Theorem: (Jordan-Chevalley) Let A be a square matrix with entries in C. Then A is similar to a Jordan matrix J.
Moreover, J is unique up to reordering the Jordan blocks.
The proof of this theorem is challenging to the extent that it isn’t found in most linear algebra textbooks.
But thinking back to the last section, we had shown that Cn decomposes into a direct sum of the eigenspaces
Vλi for the square matrix A,
C n = Vλ 1 ⊕ · · · ⊕ Vλ m ,
if and only if A is diagonalizable. If we now redefine the Vλi to be generalized eigenspaces,
Vλi = {v ∈ Cn : (A − λi In )r v = 0 for some r > 0},
then we have the following.
Corollary: Let A be an n × n matrix with entries in C, and suppose λ1 , . . . , λm are the eigenvalues of A. Then
Cn decomposes into a direct sum of the generalized eigenspaces of A,
C n = Vλ 1 ⊕ · · · ⊕ Vλ m .
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7. Algebras
Definition: An algebra A is a vector space which also has a multiplication operation, that is, an operation
which behaves in the same way as multiplication does in a field with respect to addition and scaling (e.g.,
a · 0 = 0 = 0 · a, a · (b + c) = a · b + a · c, etc. for all a, b, c ∈ A).
An algebra A is associative if multiplication is associative: (a · b) · c = a · (b · c) for all a, b, c ∈ A; it’s unital if
it has a multiplicative unit: there exists an element 1A ∈ A such that a · 1A = a = 1A · a for all a ∈ A.
Examples:
• Every field K is an associative, unital algebra as a 1-dimensional vector space over itself.
• For any vector space V, we have the endomorphism algebra End(V) of all linear maps f : V → V
(“endomorphisms” of V). Multiplication is defined to be composition of maps, g ◦ f , which is
associative. (Is it also unital?)
• The matrix algebra Mn (K) of all n × n matrices with entries in K. The multiplication is just matrix
multiplication, which is associative and unital with 1Mn (K) = In .
• The polynomial algebra K[x1 , . . . , xn ] of all polynomials in the variables x1 , . . . , xn with coefficients in K.
Multiplication is defined as usual, so K[x1 , . . . , xn ] is associative and unital.
Notice that fields K and polynomial algebras K[x1 , . . . , xn ] are commutative algebras: a · b = b · a for all a, b ∈ A.
(Is Mn (K) commutative?)
Exercise:
1. Under what circumstances is the path algebra KQ of a quiver Q unital? (They’re always associative.)
2. Write down a basis for the matrix algebra Mn (K). What’s its dimension?
3. Repeat exercise 2 for K[x1 , . . . , xn ].
Sometimes it’s convenient to define an algebra by generators and relations. It would be a royal pain to
give a completely formal definition of this, so I’ll say it informally by first giving an example.
Example: The associative, unital algebra K[x1 , . . . , xn ] is generated by the elements x1 , . . . , xn subject to the
relations xi x j = x j xi for all 1 ≤ i, j ≤ n.
In other words, every polynomial in the variables x1 , . . . , xn with coefficients in K can be gotten by
repeatedly applying the operations of addition, scaling, and multiplication starting from the generators.
(The multiplicative unit 1 was implied to exist when I used the word unital at the beginning.) On the other
hand, I still need to specify that the generators commute with each other, because otherwise, for example,
the monomials x21 x32 and x2 x1 x2 x1 x2 won’t be equal (and in this case I want them to be). Some helpful notes:
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• Different generating sets often generate different algebras, but sometimes the same one. For example,
it’s obvious that if I add another generator y to the previous example then I’ll get a different (bigger)
algebra. On the other hand, if I add something like x21 or x1 + x2 to the generating set then I’ll get the
same algebra, because those were already going to be obtained anyway so they’re redundant. For this
reason, mathematicians are often interested in minimal generating sets.
• Sometimes the equalities that make up the relations for an algebra are written as relators, elements that
are declared to be zero, instead. For example, since xi x j = x j xi is the same as xi x j − x j xi = 0, I could
have said K[x1 , . . . , xn ] has generators x1 , . . . , xn and relators xi x j − x j xi for all 1 ≤ i, j ≤ n.
• It’s debatable whether it’s better to know a set of generators and relations (a presentation) for your
algebra or to know a basis for it as a vector space, and depends on the situation. Both is probably best!
A helpful metaphor: The “game” of building up an algebra A has as its playing pieces the generators
and rules for playing them the relations.
Definition: Let A, B be algebras over the same field K. An algebra homomorphism φ : A → B is a linear map
which also preserves multiplication: φ(a1 a2 ) = φ(a1 )φ(a2 ) for all a1 , a2 ∈ A. If A, B are both unital, then φ is
required to preserve identities: φ(1A ) = 1B .
A homomorphism φ : A → B is an isomorphism if it’s also bijective. That the algebras A and B are
isomorphic is denoted by A ' B.
Exercise:
• Suppose V is a finite-dimensional vector space, dim(V) = n. Convince yourself that End(V) ' Mn (K).
• Is the map
det : Mn (K) →
M 7→
K
det M
where det M is the determinant of the matrix M an algebra homomorphism? Why or why not?
• Is the map
d
: K[x] → K[x]
dx
f (x) 7→ f 0 (x)
an algebra homomorphism? Why or why not?
• An anithomomorphism φ : A → B is a linear map which reverses multiplication: φ(a1 a2 ) = φ(a2 )φ(a1 ) for
all a1 , a2 ∈ A. Convince yourself that
(−)t : Mn (K) →
M 7→
Mn (K)
Mt ,
where Mt is the transpose of the matrix t, is an antihomomorphism.
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Recall the 3-step process we had for defining a linear map f : V → W:
1. Pick a basis {v1 , . . . , vn } of V.
2. Choose vectors f (v1 ), . . . , f (vn ) ∈ W.
3. Declare that f is obtained from these assignments by extending by linearity.
If you have algebras A, B and want an algebra homomorphism φ : A → B, there are different ways to go
about it depending on whether you want to start from having a basis of A or from having generators and
relators of A.
If you have a basis {a1 , . . . , an } of A (let’s keep things finite-dimensional for now), the process is similar
to the above:
1. Choose elements φ(a1 ), . . . , φ(an ) ∈ B.
2. Check that φ(ai · a j ) = φ(ai ) · φ(a j ) for all 1 ≤ i, j ≤ n. (Why is it enough to do this on basis vectors?)
3. Declare that φ is obtained from these assignments by extending by linearity.
If you have generators and relators for A, the process is slightly different:
1. Choose elements φ(a) ∈ B for every generator a ∈ A.
2. Check that φ(a) = 0 for every relator a ∈ A (they’re equal to zero in A and as a linear map φ(0) = 0).
3. Declare that φ is obtained from these assignments by extending by linearity and multiplicativity: φ(a · b)
is defined to be φ(a) · φ(b) for generators a, b ∈ A. (Why can’t you extend multiplicatively when starting
from a basis?)
Why exactly this last 3-step process is equivalent to defining an algebra homomorphism φ : A → B is a
deeper question. A year of graduate algebra will explain!
Definition: A representation of an algebra A is a pair (V, ρ) of a vector space V and an algebra homomorphism
ρ : A → End(V). This means for every a ∈ A there’s a linear map ρa : V → V such that
• ρa + ρb = ρa+b ,
• ρa ◦ ρb = ρab ,
so A is represented as a collection of linear transformations on a vector space.
Some associated lingo: we say A acts on V, refer to ρ as the A-action, and call V an A-module.
NOTE! For us, the terms “module” and “representation” mean exactly the same thing!
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Examples:
• A K-vector space is the same thing as a representation of the algebra K. (The action is given by scaling.)
• Every algebra A acts on itself by left-multiplication:
ρ:A
→ End(A)
ρa (b)
=
a · b.
This is called the (left) regular representation of A.
• The Weyl algebra6 W has generators x, ∂ and one relator ∂x − x∂ − 1 (so it’s almost but not quite
commutative because ∂x = x∂ + 1 , x∂). Check that the following defines an action of W on K[x].
ρ:W
→
End(K[x])
ρx ( f (x))
=
x · f (x)
ρ∂ ( f (x)
=
f 0 (x)
• Suppose W ⊆ V are vector spaces. We can find a basis of V which is adapted to W, that is, a basis
{v1 , . . . , vn } of V such that v1 , . . . , vi ∈ W and vi+1 , . . . , vn < W where i = dim W. Necessarily, {v1 , . . . , vi }
is a basis of W. In this scenario, we have a linear map called projection onto W,
πW : V → V
α1 v1 + · · · + αn vn
=
α1 v1 + · · · + αi vi .
In other words πw (vi+1 ) = · · · = πW (vn ) = 0; writing vectors of V in the adapted basis, πW just
forgets/kills all of the terms not in W. So πW restricted to W is just the identity map on W, and while
we could regard πW as going V → W or even W → W there are reasons to regard it as going V → W.
Continuing this example, let A be the associative, unital algebra with one generator z and one relator
z2 − z. Check that the following defines a representation of A.
ρ:A →
z
7→
End(V)
πW
8. Tensor Products
Definition: The tensor product V ⊗ W of two K-vector spaces V, W is the vector space consisting of all possible
linear combinations of pure tensors v ⊗ w where v ∈ V and w ∈ W. The tensor product is bilinear in the sense
• α(v ⊗ w) = αv ⊗ w = v ⊗ αw for any α ∈ K,
• (v1 + v2 ) ⊗ w = v1 ⊗ w + v2 ⊗ w, and
• v ⊗ (w1 + w2 ) = v ⊗ w1 + v ⊗ w2 .
6 Named
for Hermann Weyl (1885 - 1955), a German mathematician.
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In particular, if {v1 , . . . , vm } is a basis for V and {w1 , . . . , wn } is a basis for W, then {vi ⊗ w j : 1 ≤ i ≤ m, 1 ≤ j ≤ n}
is a basis for V ⊗ W. Consequently, dim V ⊗ W = dim V · dim W.
Exercise:
• Show that V ⊗ K ' K ' K ⊗ V by constructing explicit isomorphisms.
• Show that V ⊗ W ' W ⊗ V by constructing an explicit isomorphism.
9. Exact Sequences
Definition: For a linear map f : U → V between K-vector spaces U, V, the image of f is the set
im( f ) = {v ∈ V : v = f (u) for some u ∈ U} ⊆ V,
and the kernel (or the nullspace) of f is the set
ker( f ) = {u ∈ U : f (u) = 0} ⊆ U.
Exercise:
• Show that im( f ) is a subspace of V and ker( f ) is a subspace of U.
• Recall that f is injective (or one-to-one) if f (u1 ) = f (u2 ) implies u1 = u2 . Show that f is injective if and
only if ker( f ) = {0}.
• For use later, recall that f is surjective (or onto) if im( f ) = V.
f
g
Now suppose f : U → V and g : V → W are linear maps, so we have a sequence U →
− V→
− W.
f
g
Definition: The sequence U →
− V→
− W is called exact if ker(g) = im( f ).
f
g
Exercise: Show that if U →
− V→
− W is exact, then (g ◦ f )(u) = 0 for all u ∈ U. Is the converse true?
Definition: A (possibly infinite) sequence
fi
fi+1
· · · Vi−1 −
→ Vi −−→ Vi+1 · · ·
fi
fi+1
is called exact if every subsequence of the form Vi−1 −
→ Vi −−→ Vi+1 is exact, that is, ker( fi+1 ) = im( fi ) for all i.
f
g
Definition: A short exact sequence is an exact sequence of the form 0 → U →
− V→
− W → 0, where 0 denotes
the zero space.
Note that we don’t need any labels for the linear maps 0 → U and W → 0 because there’s only one
choice for each. Short exact sequences are of particular interest in representation theory, as we’ll see.
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f
g
Exercise: Show that 0 → U →
− V→
− W → 0 is exact if and only if f is injective and g is surjective.
f
g
Definition: A short exact sequence 0 → U →
− V→
− W → 0 splits if V ' U ⊕ W.
Exercise: Recall the following facts from linear algebra.
• The rank of a linear map f : U → V is rank( f ) = dim im( f ).
• The nullity of a linear map f : U → V is nullity( f ) = dim ker( f ).
• For every linear map f : U → V, we have dim U = rank( f ) + nullity( f ). This is the rank-nullity theorem.
Use the rank-nullity theorem to prove that every short exact sequence splits.
So, you ask, why do we have a definition for split short exact sequences if they’re all split? The reason
is that we’re about to define what it means to have a homomorphism φ : U → V between two different
representations U, V of the same algebra A, and it is absolutely not true that every short exact sequence of
them is split (the isomorphism V ' U ⊕ W would have to be as representations, not just as vector spaces).
10. Intertwiners
Recall that a representation of an algebra A is a pair (V, ρ) consisting of a vector space V and an algebra
homomorphism ρ : A → End(V). Instead of writing ρa (v) for a ∈ A and v ∈ V, let’s just write a.v and call this
“a acting on v.” So in practice, ρ is never even mentioned.
Example: The regular representation of the path algebra KQ of the quiver Q is when KQ acts on itself
(V = KQ) by left-multiplication: p.q = pq for paths p, q ∈ KQ. (Recall that pq is often zero.)
Exercise: For a path p in the quiver Q, pKQ is the set {px : x ∈ KQ}. We can define KQp similarly.
• Show that pKQ and KQp are subspaces of KQ. Are they also subalgebras?
• Describe a basis for ex KQ and a basis for Px = KQex where ex is the trivial path at the node x in Q.
Now suppose V, W are two representations of the same algebra A. So a.v for a ∈ A and v ∈ V means a
acting on v in the representation V, while a.w for w ∈ W means a acting on w in the representation W. These
could be totally different vector spaces with totally different actions.
Definition: An intertwiner φ : V → W is a linear map which also “intertwines” the two actions of A in the
sense that φ(a.v) = a.φ(v) for all a ∈ A and v ∈ V. Intertwiner is just short for homomorphism of representations.
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Example: Every linear map φ : K → K is an intertwiner of the regular representation of K. Indeed, we have
φ(α.β) = φ(αβ) = α φ(β) = α.φ(β)
for all α, β ∈ K. (If the action is scaling, then linear maps are intertwiners!)
fi
fi+1
We can now define an exact sequence · · · Vi−1 −
→ Vi −−→ Vi+1 · · · of representations of an algebra A:
It’s the same as an exact sequence of vector spaces and linear maps, except every vector space is also a
representation of A and every linear map is an intertwiner.
Definition: A representation P of an algebra A is called projective if every short exact sequence
f
g
0→U→
− V→
− P→0
of representations of A splits.
Proposition: Let Q be a quiver, K a field, and x a node in Q. Then the representation Px = KQex of the path algebra
KQ is projective.
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