In this section we look at vector spaces, and remind you of a few things that were not in the preliminaries. We also construct a few potentially new spaces. We introduce the dual space and tensor products of vector spaces for those who have not seen these. In this course all of our vector spaces will be finite-dimensional, but some of our constructions and theorems also work for infinite-dimensional spaces. In this section we will alert the reader as to some of what goes wrong with infinite-dimensional vector spaces, but our primary goal is finite-dimensional spaces and so we stick to those in the course.

The trace of a matrix is the sum of the diagonal matrices in the matrix, i.e., if a matrix \(M\) has coefficients \(m_{i,j}\) then\[\mathrm{tr}(M)=\sum_{i=1}^n m_{i,i}.\]

Lemma 1.1. Let \(A\) and \(B\) be matrices in \(M_n(k)\). We have that \(\mathrm{tr}(AB)=\mathrm{tr}(BA)\). In particular
\[ \mathrm{tr}(X^{-1}MX)=\mathrm{tr}((X^{-1}M)X)=\mathrm{tr}(X(X^{-1}M))=\mathrm{tr}(M),\]and so the trace of a matrix does not depend on the basis used to compute it.

Proof: This is easy, as we just compute the diagonal entries of \(AB\) and \(BA\). If \(A=(a_{i,j})\) and \(B=(b_{i,j})\) then
\[ \mathrm{tr}(AB)=\sum_{j=1}^n AB_{j,j}=\sum_{j,k=1}^n a_{j,k}b_{k,j}.\]The last expression is preserved by swapping \(a_{i,j}\) and \(b_{i,j}\), hence \(\mathrm{tr}(AB)=\mathrm{tr}(BA)\). □

Such properties, that do not depend on the choice of basis, are often called natural. We will see an example of something non-natural now.

We know that any two vector spaces of the same dimension are isomorphic. But if the two vector spaces are just given to you arbitrarily, there's no reason to expect there to be a natural isomorphism between them. You just pick a basis for each and send basis vectors to basis vectors; this obviously depends on the choice of basis.

Let \(V\) and \(W\) be vector spaces: define \(\mathrm{Hom}_k(V,W)\) to be the set of all vector space homomorphisms (i.e., linear transformations) from \(V\) to \(W\). This set can be endowed with a sum: if \(f_1,f_2\in \mathrm{Hom}_k(V,W)\), define \(f_1+f_2\) to be\[ v(f_1+f_2)=vf_1+vf_2.\](Here we have written \(vf\) rather than \(f(v)\). There is a good reason for this. In finite group theory maps are often written on the right, so \(fg\) is 'perform \(f\) first, then \(g\)', and \((1,2,3)=(1,2)(1,3)\). You are allowed to mix left and right structures, but it gets difficult as you often need to insert inverses in various places. Sticking to one side to write maps is the best way, even if we end up with \(vf\) rather than \(f(v)\).)
Furthermore, if \(\lambda\in k\), define \(\lambda f\) by\[ v(\lambda f)=(\lambda v)f=\lambda vf.\](Here we have put the scalar on the left. Scalars commute with everything so there's no need to be careful about this, but later in this chapter we will deal with division rings (non-commutative fields). It's important there that scalars act on the opposite side to maps.)

This turns \(\mathrm{Hom}_k(V,W)\) into a \(k\)-vector space. In the finite-dimensional case, if \(\{v_1,\dots,v_r\}\) is a basis for \(V\) and \(\{w_1,\dots,w_s\}\) is a basis for \(W\), then a basis for \(\mathrm{Hom}_k(V,W)\) is \(\phi_{i,j}\), where\[ v_i\phi_{i,j}=w_j,\qquad v_l\phi_{i,j}=0\text{ if \(l\neq i\)}.\]We see from this basis that \(\dim(\mathrm{Hom}_k(V,W))=\dim(V)\dim(W)\). In the infinite-dimensional case that formula makes no sense, but worse still, the elements of \(\mathrm{Hom}_k(V,W)\) chosen above, that map a basis element of \(V\) to a basis element of \(W\), do not form a basis for \(\mathrm{Hom}_k(V,W)\) (see Exercise 1.3).

Proposition 1.2. Let \(V\) be a \(k\)-vector space. Then there is a natural isomorphism
\[ \mathrm{Hom}_k(k,V)\cong V\]of vector spaces.

Proof: An element of \(\mathrm{Hom}_k(k,V)\) is determined by its image on \(1\). Thus we have little choice in our map: let \(\phi\) be the map taking \(v\in V\) to \(f:1\mapsto v\) in \(\mathrm{Hom}_k(k,V)\). We claim this is a \(k\)-vector space isomorphism. Since an element in \(\mathrm{Hom}_k(k,V)\) is determined uniquely by its image on \(1\), this is clearly well defined and a bijection, so all we have to prove is that is a homomorphism, but this is Exercise 1.2. □

Notice that this map is natural, in that a basis was not needed to define the map. In the next case this is not true.

Proposition 1.3. Let \(V\) be a finite-dimensional \(k\)-vector space. Then there is an isomorphism\[ \mathrm{Hom}_k(V,k)\cong V\]of vector spaces.

Proof: Let \(\{v_1,\dots,v_r\}\) be a basis for \(V\). A basis for \(\mathrm{Hom}_k(V,k)\) can be given as \(f_i:V\to k\) for which \(f_i:v_i\to 1\) and \(f_i:v_j\to 0\) for all \(j\neq i\). Exercise 1.1 is that this is a basis for \(\mathrm{Hom}_k(V,k)\). □

Notice that this isomorphism is not natural, it depends on a basis, and if \(V\) is not finite-dimensional then the two spaces need not be isomorphic. The space \(\mathrm{Hom}_k(V,k)\) is called the dual space of \(V\), denoted by \(V^*\). It is sometimes called the space of linear functionals on \(V\). If \(V\) is infinite-dimensional then \(V^*\) has strictly larger dimension than \(V\); see Exercise 1.1. That exercise also shows that \(V^{**}\) is naturally isomorphic to \(V\), at least for finite-dimensional vector spaces.

Naturality is not important, in some sense, because the two things are still isomorphic. Naturality is important though for two reasons. First, if two objects are naturally isomorphic, and we place some extra mathematical structure on top of one of them, we might well expect that isomorphism to carry that structure onto a 'reasonable' structure for the other object. This is a philosophical reason, rather than a purely mathematical one, as it more or less states 'natural isomorphisms often send nice concepts to nice concepts'.

The second reason is pedagogical. People often don't like the idea of an unnatural isomorphism. It's difficult to understand why a (finite-dimensional) vector space is isomorphic to its dual, and the dual space in general isn't so easy to comprehend.

The first reason is most important for us. We are going to be looking at representations of groups, i.e., homomorphisms from a group to \(\mathrm{GL}(V)\) for some vector space \(V\). Natural isomorphisms between vector spaces are much more likely, and indeed will, preserve all of these group actions on vector spaces, whereas non-natural isomorphisms might well not.

We also recall the construction of the direct sum. If \(V\) and \(W\) are vector spaces, with bases \(\{v_1,\dots,v_r\}\) and \(\{w_1,\dots,w_s\}\), then the vector space \(V\oplus W\), called the direct sum, has basis \(\{v_1,\dots,v_r,w_1,\dots,w_s\}\). This looks like it depends on a basis, but it can also be written as\[ V\oplus W=\{(v,w)\mid v\in V,w\in W\},\]which turns it into a natural construction. The direct sum is meant to look like addition: the dimension of \(V\oplus W\) is simply \(\dim(V)+\dim(W)\), for instance. This also works for infinite-dimensional vector spaces.

We will provide another way of taking two vector spaces and producing a third, analogous to multiplication, called the tensor product, and written \(V\otimes W\). This time \(\dim(V\otimes W)=\dim(V)\dim(W)\), so it is analogous.

If \(V\) and \(W\) have bases \(\{v_1,\dots,v_r\}\) and \(\{w_1,\dots,w_s\}\) respectively, then \(V\otimes W\) is the vector space with basis the symbols \(v_i\otimes w_j\), for \(v_i\) a basis element of \(V\) and \(w_j\) a basis element of \(W\). An arbitrary element of \(V\otimes W\) looks like\[ \sum_{i,j} \lambda_{i,j} v_i\otimes w_j,\]where \(\lambda_{i,j}\in k\).

Again, this doesn't look natural, but we can make it look natural by defining it as being given by all finite sums\[ \sum v\otimes w,\]for various \(v\in V\) and \(w\in W\). We need to be careful, because with that definition, \((v_1+v_2)\otimes w\) and \((v_1\otimes w)+(v_2\otimes w)\) look like different elements. Formally we need to quotient out by these relations, but informally we say that we force the distributive law to hold in both variables. We also need to make sure that \(\lambda v\otimes w\) and \(v\otimes \lambda w\) are the same thing, so we further need to impose the relations \(\lambda v\otimes w=\lambda(v\otimes w)\) and \(v\otimes \lambda w=\lambda(v\otimes w)\).

With this forcing of the distributive and scalar laws, this becomes a space with basis as defined above, and so the space \(V\otimes W\) is a natural construction as well. Again, we can do this for infinite-dimensional spaces.

We have introduced three new types of vector spaces: \(\mathrm{Hom}_k(V,W)\), \(V^*\) and \(V\otimes W\). We now get to play with all of them by proving an isomorphism including all three. We will prove that the spaces \(\mathrm{Hom}_k(V,W)\) and \(V^*\otimes W\) are isomorphic. Well, of course they are, because they have the same dimension: \(\dim(V)\dim(W)\). But we want to prove a natural isomorphism. The next result is related to tensor-Hom adjunction, which is the statement that\[ \mathrm{Hom}(A\otimes B,C)\cong \mathrm{Hom}(A,\mathrm{Hom}(B,C)).\]This holds for all \(k\)-vector spaces, see Exercise 1.13. The cleaner version we prove here holds only for finite-dimensional vector spaces, but in that situation implies tensor-Hom adjunction. We will call this version tensor-Hom adjunction, because for us this is exactly what we need.

Proposition 1.4. Let \(V\) and \(W\) be finite-dimensional \(k\)-vector spaces. There is a natural isomorphism\[ V^*\otimes W\to\mathrm{Hom}_k(V,W)\]given by \(f\otimes w\) being sent to \(f_w:v\mapsto (vf)\cdot w\).

Proof: First we want to check that the map we given makes sense, so fix \(f\in V^*\), \(w\in W\) and \(v\in V\). The element \(f\) is a linear transformation from \(V\) to \(k\), so \(vf\) is simply a scalar \(\lambda\in k\). As \(W\) is a vector space, \((vf)\cdot w\) makes sense, and is an element of \(W\), so \(f_w\) really is a function from \(V\) to \(W\). We have to check that this is a linear transformation, but if \(v_1,v_2\in V\) and \(\mu\in k\), then\[ (v_1+\mu v_2)f_w=(v_1+\mu v_2)f\cdot w=(v_1f)\cdot w+\mu(v_2f)\cdot w=v_1f_w+\mu v_2f_w.\]This proves that \(f_w\) is not just a function from \(V\) to \(W\), but an element of \(\mathrm{Hom}_k(V,W)\).

We also have only defined the map on some elements of \(V^*\otimes W\), and we claim this is a map on all of \(V^*\otimes W\). Clearly, since the set \(f\otimes w\) spans \(V^*\otimes W\), we can extend the map by linearity, but we need to be sure that this is well defined, i.e., if \(f_1,f_2\in V^*\) and \(\mu\in k\) then \((f_1+\mu f_2)\otimes w\) and \((f_1\otimes w)+\mu(f_2\otimes w)\) get sent to the same image, and we must do the same with \(f\otimes(w_1+\mu w_2)\) and \(f\otimes w_1+\mu(f\otimes w_2)\). This is a very similar calculation to that above:
\begin{align*}v(f_1+\mu f_2)_w&=(v(f_1+\mu f_2))\cdot w
\\ &=(vf_1+\mu vf_2)\cdot w
\\ &=(vf_1)\cdot w+(\mu vf_2)\cdot w
\\ &=v(f_1)_w+\mu v(f_2)_w
\end{align*}So \((f_1+\mu f_2)\otimes w=v\otimes f_1+\mu(v\otimes f_2)\).

We won't prove that the same distributivity property happens on the right-hand side, but if you are still feeling a little uncomfortable with all these maps and symbols flying about and being nested inside other functions then you should do so.

Thus we have proved that the map \(F:f\otimes w\mapsto f_w\) has the correct codomain. We need to show that it is a linear transformation as well, but that is more or less exactly the same computation as above, when we showed that it was well defined, just a mixture of the two computations. We won't do it here, but you can check it yourself.

What we will show is that this map is not just a linear transformation but an isomorphism of vector spaces. To show that \(F\) is surjective is easy: fixing bases \(\{v_i\}\) and \(\{w_j\}\) for \(V\) and \(W\), we also get bases \(\{f_i\}\) for \(V^*\) and \(\phi_{i,j}\) for \(\mathrm{Hom}_k(V,W)\), where \(v_l\phi_{i,j}=\delta_{i,l}w_j\). If we can find an element of \(V^*\otimes W\) whose image is \(\phi_{i,j}\) then we are done, but this is \(f_i\otimes w_j\):\[ v_l(f_i)_{w_j}=(v_lf_i)\cdot w_j=v_l\delta_{i,l}w_j=v_l\phi_{i,j},\]as needed.

To show directly that \(F\) is injective is more fiddly, but since we know the dimensions are equal and \(F\) is a linear transformation, it must be an isomorphism. □

Using this result, we can turn any collection of Hom-spaces into a tensor product of vector spaces and their duals via a natural map. We just need one more fact, which is that \((V\otimes W)^*\) is naturally isomorphic to \(V^*\otimes W^*\), see Exercise 1.12 (and to recall that \(V^{**}\) is naturally isomorphic to \(V\) by Exercise 1.1). For example, to prove the general form of tensor-Hom adjunction we have\[ \mathrm{Hom}(A\otimes B,C)\cong (A\otimes B)^*\otimes C\cong A^*\otimes B^*\otimes C,\]and\[ \mathrm{Hom}(A,\mathrm{Hom}(B,C))\cong \mathrm{Hom}(A,B^*\otimes C)\cong A^*\otimes B^*\otimes C.\]Thus the two sides are naturally isomorphic.

If \(V\) is finite-dimensional but \(W\) is infinite-dimensional then \(\mathrm{Hom}_k(V,W)\) is still naturally isomorphic to \(V^*\otimes W\). If \(V\) is infinite-dimensional we do not get an isomorphism, but \(V^*\otimes W\) maps to the finite-rank maps in \(\mathrm{Hom}_k(V,W)\). One can see why we make a standing assumption that our vector spaces are finite-dimensional.