## 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 to appropriate categories of -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 to be the “free” graded-commutative algebra on the vector space . That’s a tip-off that we’re thinking should be the left adjoint of the “forgetful” functor ~~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 ) 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 which we’re going to make the right-adjoint. Now for each vector space we need a graded-commutative algebra — clearly the one we’ll pick is — and a universal arrow . The underlying vector space of the exterior algebra is the direct sum of all the spaces of antisymmetric tensors on .

Yesterday we wrote this without the , since we often just omit forgetful functors, but today we want to remember that we’re using it. But we know that , so the obvious map to use is the one that sends a vector to itself, now considered as an antisymmetric tensor with a single tensorand.

But is this a *universal* arrow? That is, if is another graded-commutative algebra, and is another linear map, then is there a unique homomorphism of graded-commutative algebras so that ? Well, tells us where in we have to send any antisymmetric tensor with one tensorand. Any other element in is the sum of a bunch of terms, each of which is the wedge of a bunch of elements of . So in order for to be a homomorphism of graded-commutative algebras, it has to act by simply changing each element of in our expression for into the corresponding element of , 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 . And this establishes that is the object-function of a functor which is left-adjoint to .

So how does work on morphisms? It’s right in the proof above! If we have a linear map , we need to find some homomorphism . But we can compose with the linear map , which gives us . The universality property we just proved shows that we have a unique homomorphism . And, specifically, it is defined on an element by writing down in terms of vectors in and applying to each vector in the expression to get a sum of wedges of elements of , which will be an element of the algebra .

Of course, as stated above, we get similar constructions for the commutative algebra and the tensor algebra .

Since, given a linear map the induced homomorphisms , , and preserve the respective gradings, they can be broken into one linear map for each degree. And if is invertible, so must be its image under each functor. These give exactly the tensor, symmetric, and antisymmetric representations of the group , if we consider how these functors act on invertible morphisms . Functoriality is certainly a useful property.

I think there’s something wrong with the statement of the universal property. First let me give an example, and then what I think is the right statement.

Let $V$ be a 1-dimensional space spanned by an element $v$, and let $A$ be the graded commutative algebra generated by a single element $x$ in degree 2 (so as an algebra $A$ is a polynomial algebra; the grading is concentrated in even degrees). The total space $U(A)$ is just the vector space of polynomials in $x$, and we may define a linear map $V \to U(A)$ sending $v$ to $x$. But there is no graded algebra map $\Lambda(V) \to A$, because the square of $v$ in $\Lambda(V)$ is 0 and the square of $x$ in $A$ is not.

It seems to me you want to say this: let $U$ be the functor from graded algebras to spaces which sends a graded algebra $A$ to the space $A_1$ of degree 1 elements. Then $\Lambda$ is left adjoint to $U$. You can tell because you defined the grading on $\Lambda(V)$ so that $\Lambda^m(V)$ is the degree $m$ component, so that the universal map $V \to U(\Lambda(V))$, mapping $V$ onto the part $\Lambda^1(V)$, interprets elements of $V$ as degree 1 elements.

Comment by Todd Trimble | October 28, 2009 |

Todd, just thought of something while up sleepless: what about the category of graded-commutative algebras and maps that respect the grading. That’d probably work better.

Comment by John Armstrong | October 29, 2009 |

[...] As a symmetric bilinear form, the inner product provides us with an isomorphism . Now we can use functoriality to see what this does for our tensor algebras. Again, I’ll be mostly interested in the [...]

Pingback by Tensor Algebras and Inner Products I « The Unapologetic Mathematician | October 29, 2009 |

The morphisms in the category I was referring to, the category of graded-commutative algebras, are assumed to respect the grading already. So in the example I gave, there is just one graded algebra map from $\Lambda(V)$ to $A$; it sends everything in $\Lambda^1(V)$ to the zero element in $A_1$, which is the only element $A_1$ has. [I didn't say it this way before, but I should have.] But on the other hand, there are lots of linear maps from $V$ to the total space of $A$ (the $U(A)$ you had prior to the edit).

So right, I was saying that for general vector spaces $V$ and graded-commutative algebras $A$, linear maps $V \to A_1$ are in natural bijection with morphisms of graded-commutative algebras $\Lambda(V) \to A$. If that’s what you’re saying now, then we’re in agreement.

Comment by Todd Trimble | October 29, 2009 |

[...] linear transformation sends to the vector . By functoriality, it sends to . And now we want to calculate the [...]

Pingback by Inner Products on Exterior Algebras and Determinants « The Unapologetic Mathematician | October 30, 2009 |