The dual of a tensor algebra as a right module

Question

Let $V$ be a finite-dimensional vectorspace over a field $k$, and let $T$ denote the tensor algebra of $V$ (thought as a graded $k$-algebra). Denote by $T^\vee$ the dual of $T$, i.e. $\mathrm{Hom}_k(T,k)$. $T^\vee$ is a graded right $T$-module where the action is defined as follows: if $\phi\in T^\vee$ and $a,b\in T$ then $(\phi a)(b)=\phi(ab)$. Do you know of a nice generating set of $T^\vee$ as a graded right $T$-module? (minimal generating set would be the best)

Answer 1

Here's how I would imagine this module:

Fix a basis of $V$. This essentially identifies the tensor algebra with a polynomial ring in noncommuting variables. The dual can be taken to have the same underlying set. The monomial $x_{i_1}x_{i_2}\cdots x_{i_k}$ in the dual satisfies $$x_{i_1}x_{i_2}\cdots x_{i_k}(x_{\lambda})=\delta_{\lambda,(i_k,i_{k-1},\ldots,i_2,i_1)}$$ That is, the element of the dual basis corresponding to a monomial is the same monomial in reverse. The action of $T$ is then given by $$x_{i_1}x_{i_2}\cdots x_{i_k}\cdot x_j=\left\{\begin{array}{ll} 0&\mbox{ if }i_k\neq j\\x_{i_1}x_{i_2}\cdots x_{i_{k-1}}&\mbox{ if }i_k=j\end{array}\right.$$ That is, $T$ acts by right cancellation. A minimal generating set would be given by a set of monomials $x_{\lambda_i}$ for compositions $\lambda_i$ with $i$ in some index set $I$ where

1. Every composition representing a monomial is a prefix of $\lambda_i$ for some $i\in I$.
2. If $i\neq j$ are in $I$, then $\lambda_i$ is not a prefix of $\lambda_j$.

For example, the composition $(1,3,2)$ corresponds to the monomial $x_1x_3x_2$, and $(1,3)$ is a prefix of $(1,3,2)$.