# How obviously injective is this “graded symmetrizer” map $\operatorname{S}(V) \to \operatorname{T}(V)$?

Starting with a graded vector space $V$, you can construct the tensor algebra $\operatorname{T}(V) := \bigoplus_{n>0} V^{\otimes n}$ and you can construct the symmetric algebra $\operatorname{S}(V) := \operatorname{T}(V)/\langle v \otimes w - (-1)^{|v||w|}w \otimes v \mid v,w \in V \rangle$. Now consider the map $\chi \colon \operatorname{S}(V) \to \operatorname{T}(V)$ where, for an element in the $n^\text{th}$ graded component of $\operatorname{S}(V)$, $$\chi \colon (v_1, \dotsc, v_n) \mapsto \sum_{\sigma \in S_n} \epsilon(\sigma) \left(v_{\sigma(1)} \otimes \dotsb \otimes v_{\sigma(n)} \right) \,,$$ where $\epsilon(\sigma)$, the total Koszul sign, is just a sign we get from the sign of $\sigma$ and the cost of switching $v_i$ and $v_j$ in the tensor. According to the notes I'm reading here and here, this map is injective. But how obvious is this fact? Is it enough to say that $\chi$ is a (multiple of a) section of the quotient map $\operatorname{T}(V) \to \operatorname{S}(V)$, so it must be injective?

• Also, note these related questions here and here (so they'll show up on the sidebar). – Mike Pierce Dec 13 '17 at 17:27
• Wouldn't $\epsilon$ make it alternating? – Andres Mejia Dec 13 '17 at 17:49
• @AndresMejia Yeah, I think worse even, depending on the degrees of the $v_i$. So it isn't immediately a multiple of a section, but it's only kinda almost like a section. ... – Mike Pierce Dec 13 '17 at 17:53
• @AndresMejia: No. His $\epsilon$ is the Koszul sign, not the sign of the permutation. – darij grinberg Dec 13 '17 at 18:07
• @darijgrinberg thanks – Andres Mejia Dec 13 '17 at 18:07

From those notes, $\epsilon(\sigma)$ is the sign satisfying $\epsilon(\sigma)v_{\sigma(1)}\ldots v_{\sigma(n)}=v_1\ldots v_n$. Then if $\pi:T(V)\rightarrow S(V)$ is the projection, for any $v_1v_2\ldots v_n\in S^n(V)$, $$\pi\chi(v_1\ldots v_n)=\sum \epsilon(\sigma)\pi(v_{\sigma(1)}\otimes \ldots\otimes v_{\sigma(n)})=\sum \epsilon(\sigma)v_{\sigma(1)}\ldots v_{\sigma(n)}=\sum v_1\ldots v_n=(n!)v_1\ldots v_n$$ so $\pi\chi$ is injective on $S^n(V)$ (in characteristic $0$), hence injective on $S(V)$. If the composition is injective, then $\chi$ is necessarily injective.