# The Unapologetic Mathematician

## Functoriality of Tensor Algebras

The three constructions we’ve just shown — the tensor, symmetric tensor, and exterior algebras — were all asserted to be the “free” constructions. This makes them functors from the category of vector spaces over $\mathbb{F}$ to appropriate categories of $\mathbb{F}$-algebras, and that means that they behave very nicely as we transform vector spaces, and we can even describe exactly how nicely with explicit algebra homomorphisms. I’ll work through this for the exterior algebra, since that’s the one I’m most interested in, but the others are very similar.

Okay, we want the exterior algebra $\Lambda(V)$ to be the “free” graded-commutative algebra on the vector space $V$. That’s a tip-off that we’re thinking $\Lambda$ should be the left adjoint of the “forgetful” functor $U$ which sends a graded-commutative algebra to its underlying vector space (Todd makes a correction to which forgetful functor we’re using below). We’ll define this adjunction by finding a collection of universal arrows, which (along with the forgetful functor $U$) is one of the many ways we listed to specify an adjunction.

So let’s run down the checklist. We’ve got the forgetful functor $U$ which we’re going to make the right-adjoint. Now for each vector space $V$ we need a graded-commutative algebra — clearly the one we’ll pick is $\Lambda(V)$ — and a universal arrow $\eta_V:V\rightarrow U(\Lambda(V))$. The underlying vector space of the exterior algebra is the direct sum of all the spaces of antisymmetric tensors on $V$.

$\displaystyle U(\Lambda(V))=\bigoplus\limits_{n=0}^\infty A^n(V)$

Yesterday we wrote this without the $U$, since we often just omit forgetful functors, but today we want to remember that we’re using it. But we know that $A^1(V)=V$, so the obvious map $\eta_V$ to use is the one that sends a vector $v$ to itself, now considered as an antisymmetric tensor with a single tensorand.

But is this a universal arrow? That is, if $A$ is another graded-commutative algebra, and $\phi:V\rightarrow U(A)$ is another linear map, then is there a unique homomorphism of graded-commutative algebras $\bar{\phi}:\Lambda(V)\rightarrow A$ so that $\phi=U(\bar{\phi})$? Well, $\phi$ tells us where in $A$ we have to send any antisymmetric tensor with one tensorand. Any other element $\upsilon$ in $\Lambda(V)$ is the sum of a bunch of terms, each of which is the wedge of a bunch of elements of $V$. So in order for $\bar{\phi}$ to be a homomorphism of graded-commutative algebras, it has to act by simply changing each element of $V$ in our expression for $\upsilon$ into the corresponding element of $A$, and then wedging and summing these together as before. Just write out the exterior algebra element all the way down in terms of vectors, and transform each vector in the expression. This will give us the only possible such homomorphism $\bar{\phi}$. And this establishes that $\Lambda(V)$ is the object-function of a functor which is left-adjoint to $U$.

So how does $\Lambda$ work on morphisms? It’s right in the proof above! If we have a linear map $f:V\rightarrow W$, we need to find some homomorphism $\Lambda(f):\Lambda(V)\rightarrow\Lambda(W)$. But we can compose $f$ with the linear map $\eta_W$, which gives us $\eta_W\circ f:V\rightarrow U(\Lambda(W))$. The universality property we just proved shows that we have a unique homomorphism $\Lambda(f)=\overline{\eta_W\circ f}:\Lambda(V)\rightarrow\Lambda(W)$. And, specifically, it is defined on an element $\upsilon\in\Lambda(V)$ by writing down $\upsilon$ in terms of vectors in $V$ and applying $f$ to each vector in the expression to get a sum of wedges of elements of $W$, which will be an element of the algebra $\Lambda(W)$.

Of course, as stated above, we get similar constructions for the commutative algebra $S(V)$ and the tensor algebra $T(V)$.

Since, given a linear map $f$ the induced homomorphisms $\Lambda(f)$, $S(f)$, and $T(f)$ preserve the respective gradings, they can be broken into one linear map for each degree. And if $f$ is invertible, so must be its image under each functor. These give exactly the tensor, symmetric, and antisymmetric representations of the group $\mathrm{GL}(V)$, if we consider how these functors act on invertible morphisms $f:V\rightarrow V$. Functoriality is certainly a useful property.

October 28, 2009