Multilinear algebra

5 Traces and TQFT
Another way that Hom(V, W ) is related to tensor products is the following. Given f ∈ Hom(V, W ) and v ∈ V ,
we can evaluate f on v to get an element f(v) ∈ W . This is a bilinear map Hom(V, W )×V → W , so it gives a
linear map ev : Hom(V, W )⊗V → W . In particular, for W = k, we get a map tr : V ⊗V ∗ → k. Furthermore,
if we identify Hom(V, W ) with V ∗ ⊗W = W ⊗V ∗ as in Theorem 4.5, ev : Hom(V, W )⊗V = W ⊗V ∗ ⊗V → W
is just given by pairing up the V ∗ and V using tr. That is ev(w ⊗ v ∗ ⊗ v) = v ∗ (v)w (for v ∗ ∈ V ∗ ). This can
be seen by the map used in Theorem 4.5: an element of W ⊗ V ∗ maps V to W by pairing the element of v
with the V ∗ part of the tensor product.
More generally, there is a composition map Hom(U, V ) ⊗ Hom(V, W ) → Hom(U, W ) which composes two
linear maps. Writing these Hom’s in terms of tensor products and duals, this is a map U ∗ ⊗ V ⊗ V ∗ ⊗ W →
U ∗ ⊗ W . A similar analysis to the analysis above shows that this map is just given by pairing up the V and
V ∗ in the middle: we compose linear maps by sending u ∗ ⊗ v ⊗ v ∗ ⊗ w to v ∗ (v)u ∗ ⊗ w.
Let’s now see what happens when we let W = V . We then have Hom(V, V ) = V ⊗ V ∗ , and there is a
map tr : Hom(V, V ) = V ⊗ V ∗ → k, a map which takes a linear map T : V → V and gives a scalar. This is
called the trace tr(T ) of T .
You may have seen the trace of a matrix defined as the sum of its diagonal entries. This definition is really
quite mystifying: why on earth would you take the diagonal entries (as opposed to some other collection of
entries) and add them up? Why on earth would that be independent of a choice of basis?
On the other hand, our definition is manifestly independent of a choice of basis and is very naturally
defined: it’s just the natural evaluation map on V ⊗ V ∗ . We can furthermore check that this agrees with
the “sum of the diagonal entries” definition.
Example 5.1. Let V have basis {e1, e2} and let {α 1 , α 2 } be the dual basis of V ∗ . Let T : V → V have
a b
matrix
c d
matrix with ij entry 1 and all other entries 0. Thus T is given by
. We want to write down T as an element of V ⊗ V ∗ . Recall that ei ⊗ α j corresponds to the
ae1 ⊗ α 1 + be1 ⊗ α 2 + ce2 ⊗ α 1 + de2 ⊗ α 2 .
Now the trace just takes these tensors and evaluates them together, so ei ⊗ α j goes to 1 if i = j and 0
otherwise. Thus tr(T ) ends up being exactly a + d, the sum of the diagonal entries. The same argument
would generalize to n × n matrices for any n.
Besides basis-invariance, one of the most important properties of traces is the following.
Theorem 5.2. Let S, T : V → V be linear maps. Then tr(ST ) = tr(T S).
Proof. Write S and T as elements of V ∗ ⊗V , say S = v ∗ ⊗v and T = w ∗ ⊗w (actually, S and T will be linear
combinations of things of this form, but by linearity of everything we can treat each term separately). Then
recall that we compose linear maps by “pairing up the vectors in the middle”, so ST will be v ∗ (w)w ∗ ⊗ v, so
tr(ST ) = v ∗ (w)w ∗ (v). On the other hand, T S = w ∗ (v)v ∗ ⊗ w and tr(T S) = w ∗ (v)v ∗ (w). But multiplication
of scalars is commutative, so these are the same!
Let’s now look more closely at how duals and tensor products interact with linear maps. First, note that
an element u of a vector space U is just the same thing as a linear map T : k → U, namely the linear map
such that T (1) = u. Thus in showing that elements of V ∗ ⊗ W = Hom(V, W ) are the same as maps V → W ,
we’ve shown that maps V → W are the same as maps k → V ∗ ⊗ W . We can think of this as “moving V to
the other side” by dualizing it. In fact, this works more generally.
Theorem 5.3. Let U, V , and W be (finite-dimensional) vector spaces. Then linear maps U ⊗ V → W are
naturally in bijection with linear maps U → V ∗ ⊗ W = Hom(V, W ).
Proof. Given T : U ⊗ V → W , we can define ˜ T : U → Hom(V, W ) by ˜ (T )(u)(v) = T (u ⊗ v). Conversely,
given ˜ T : U → Hom(V, W ), we can define T : U ⊗ V → W by T (u ⊗ v) = ˜ T (u)(v). Clearly these two
operations are inverse to each other.
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