In this old paper, Harrison develops what is today known as Harrison cohomology. On page 192 Harrison writes:
Let $k$ be a field and $A$ a commutative associative algebra over $k$. Let $T$ be the tensor algebra (without the usual identity adjoined) of $A$. Then $T$ with the shuffle product is a skew-commutative, graded algebra.
I read the passage as follows:
We define $T:=\bigoplus_{n\geq 1}A^{\otimes n}.$ (The sentence "without the usual identity adjoined" seems to mean that we consider $n\geq 1$ and not $n\geq 0$.) The shuffle product on $T$ is normally defined as the $k$-linear map $\mathrm{shf}:T\otimes T\to T$ by $\mathrm{shf}\left(\left(a_1\otimes a_2\otimes ...\otimes a_i\right)\otimes\left(a_{i+1}\otimes a_{i+2}\otimes ...\otimes a_n\right)\right)= \sum\limits_{\sigma\in\mathrm{Sh}\left(i,n-i\right)} a_{\sigma^{-1}\left(1\right)} \otimes a_{\sigma^{-1}\left(2\right)} \otimes ... \otimes a_{\sigma^{-1}\left(n\right)}$ for every $n\in \mathbb N$ and $a_1,a_2,...,a_n\in V$. Here, $\mathrm{Sh}\left(i,n-i\right)$ denotes the set of all $\left(i,n-i\right)$-shuffles, i. e. of all permutations $\sigma\in S_n$ satisfying $\sigma\left(1\right) < \sigma\left(2\right) < ... < \sigma\left(i\right)$ and $\sigma\left(i+1\right) < \sigma\left(i+2\right) < ... < \sigma\left(n\right)$.
If I read the passage correctly, I am confused. Isn't the shuffle product commutative? Does Harrison mean something else by the shuffle product, that is consider a different, now skew-commutative, product on $T$?