Proving the graded structure of the tensor algebra from only the universal property

Question

When the tensor algebra is presented, it is usually constructed as the direct sum of all tensor powers of the space. By this construction, the graded structure of the tensor algebra is easy to prove. However, I have not been able to find any proofs of the graded structure of the tensor algebra using just the universal property. Here's my question: Has anybody come up with a proof that the tensor algebra is graded using just the universal property? I will admit that I have a hard time using universal properties in general to prove things about objects, so I've had a hard time trying to come up with an argument myself.

The reason I'm interested in this is that I've started studying category theory recently, and I've heard about the adjoint functor theorem, which seems to provide a way to define the tensor algebra without using the traditional construction in the first place (I may be wrong about this since I haven't learned enough category theory yet). I was wondering how easy it would be to prove that the tensor algebra is graded from this kind of definition.

Another reason I'm interested in this is that from studying geometric algebra, I've seen that there is an alternative grading structure for Clifford algebras that is similar to the grading of the tensor algebra (that only respects anti-commuting products, not arbitrary products like is usually implied by the use of the word "grade"), and I was wondering if an argument for the grading of the tensor algebra could be adapted to proving this as well. I'm not expecting an answer to this problem though; the question is about tensor algebras.

Answer 1

$\newcommand\T{\mathrm T}$I assume that "the universal property" means that, for every $R$-algebra $S$, every $R$-module map $V \to S$ extends uniquely to an $R$-algebra map $\T V \to S$.

In the original version of this answer, I proceeded as follows. For each $v^* \in V^* \mathrel{:=} \operatorname{Hom}_\text{\(R\)-mod}(V, R)$, let $\alpha_{v^*}$ be the unique extension of the map $V \to R[x]$ given by $v \mapsto \langle v^*, v\rangle x$ to an $R$-module map $\T V \to R[x]$. Then, for each $n \in \mathbb Z_{\ge 0}$, $\T^nV$ is the set of $\tau \in \T V$ such that $\alpha_{v^*}(\tau)$ lies in $R x^n$ for all $v^* \in V^*$.

This definition is too coarse: it correctly identifies the degree of each homogeneous element, but, as you pointed out, it can also incorrectly assign elements degrees that they shouldn't have.

The obvious fix is to replace $R$-module maps $V \to R$ by $R$-module maps $V \to A$ for arbitrary $R$-algebras $A$. This works, but the only way that I can see to show that it always works is to take $A$ to be the usual explicitly constructed tensor algebra, or something close to it. For example, I have already cheated a bit by using $R[x]$, which is just the tensor algebra on $R^*$; but I could cheat even more by taking for $A$ the quotient of the polynomial $R$-algebra in $V$-many indeterminates by $r x_v = x_{r v}$. Then there is a natural homomorphism $V \to A$ given by $v \mapsto x_v$, and we could use the above trick on $\T V \to A[x]$, or just observe that $A$ itself is graded (since we're quotienting out by a homogeneous ideal) and use the degree there. But at that point we've practically constructed the tensor algebra.

Answer 2

LSpice asked me to expand my old comment into an answer.

Let me write $R$ for whatever base commutative ring you are working over, $V$ for your $R$-module, and $TV := T_R V$ for its "tensor algebra", defined by: for any (not necessarily commutative) $R$-algebra $A$, maps $T_R V \to A$ of $R$-algebras are the same as maps $V \to A$ of $R$-modules. In other words, $T_R$ is the left adjunct to the forgetful map $\{R$-algebras$\} \to \{R$-modules$\}$. Since $T_R$ is a functor, and since the group $R^\times$ of invertible elements of $R$ acts by rescaling on $V$, we get automatically a natural action of $R^\times$ on $T_R V$.

Now for any commutative $R$-algebra $S$, I claim that there is a natural equality $$ TV \otimes S := T_S (V \otimes_R S) \cong (T_R V) \otimes_R S.$$ Indeed, to show this I simply need to show that the RHS satisfies the correct universal property for the LHS, i.e. test it against maps into $S$-algebras $A$. But if $B$ is any $R$-algebra, and $A$ is any $S$-algebra, then the hom-tensor adjunction identifies $R$-algebra maps $B \to A$ with $S$-algebra map $B \otimes_R S \to A$. The claim follows.

But $S^\times$ acts on $TV \otimes S$, extending the $R^\times$-action from before. These actions compile into an action on $TV$ by the affine algebraic group $\mathbb{G}_m = \mathrm{Spec}_R(R[z^{\pm 1}]) : S \mapsto S^\times$.

Now, I claim that for any $R$-module $W$, an action on $W$ by $\mathbb{G}_m$ is the same data as a $\mathbb{Z}$-grading of $W$. But this is just the Peter–Weyl theorem and the fact that the regular representation $\mathcal{O}(\mathbb{G}_m) = R[z^{\pm 1}]$ breaks up as a direct sum of $\mathbb{Z}$-many one-dimensional modules.

In detail, because we are working with algebraic groups, it is well-defined to talk about "the subspace $W_n$ of $W$ on which $z \in \mathbb{G}_m$ acts by $z^n$", and it is easy to show that this is a direct summand. The only slightly non-easy fact is that the inclusion $\bigoplus_n W_n \subset W$ is an isomorphism, or in other words that every element of $W$ is a finite sum of elements with well-defined eigenvalues. The cleanest way to show this is to convince yourself that an action of an affine algebraic group $G$ on a module $W$ consists of a coaction $W \to \mathcal{O}(G) \otimes W$; then the statement that we need (about decomposing as a finite sum) follows just the fact that elements of any tensor product, and in particular $\mathcal{O}(G) \otimes W$, are finite sums of pure tensors.

Answer 3

After all of this time I found a simple and direct solution. I can't believe I didn't think of this before:

Let $V$ be an $R$-Module and let $T(V)$ be its tensor algebra. Define $T^n(V)$ to be the span of all products of $n$ vectors in $T(V)$, and consider the (for now possibly distinct) space $\bigoplus_{n = 0}^\infty T^n(V)$. Multiplying two homogeneous tensors in this direct sum produces another homogeneous tensor, and we can extend this by linearity to show that $\bigoplus_{n = 0}^\infty T^n(V)$ is a graded algebra.

Now let $A$ be an arbitrary algebra with a module homomorphism $f : V \to A$. By the universal property of the tensor algebra, this extends to an algebra homomorphism $g : T(V) \to A$. We can define the restriction of $g$ onto each $T^n(V)$ and then extend by linearity to produce a map $h : \bigoplus_{n = 0}^\infty T^n(V) \to A$. This map is an algebra homomorphism that is an extension of $f$ (and it's trivial to show that this extension is unique), showing that $\bigoplus_{n = 0}^\infty T^n(V)$ satisfies the universal property that uniquely defines the tensor algebra. Thus, $T(V) \cong \bigoplus_{n = 0}^\infty T^n(V)$, showing that $T(V)$ itself is graded.