$n$-th Symmetric tensors as subspace of $n$-th tensor

Question

Let $V$ be a finite dimensional vector space over $\mathbb{C}$.

Let $T^n(V)=V\otimes \cdots \otimes V$ ($n$-times).

Let $S_n'(V)$ be the subspace of $T^n(V)$ spanned by $$(*)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{n!} \sum_{\sigma} v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(n)} \,\,\,\,( v_i\in V).$$ Let $I$ be the ideal of $T(V)=\mathbb{C}\oplus V\oplus (V\otimes V) \oplus \cdots =\bigoplus_{n=0}^{\infty} T^n(V)$ generated by vectors $x\otimes y-y\otimes x$ for all $x,y\in V$. Then

$S_n'(V)$ is a complement of $T^n(V)\cap I$ in $T^n(V)$.

Q. When stating this fact in Bourbaki's Lie groups and Lie Algebra Chapter 1-3 (P. 16) or in Knapp's Basic Algebra (P. 287), I didn't understand what is the role of taking $\frac{1}{n!}$? The subspace spanned by vectors $(*)$ is same as that spanned by removing the coefficient $\frac{1}{n!}$ in $(*)$. What is the significance of term $\frac{1}{n!}$?

Answer 1

The reason why it is stated that way is that $$ \pi\colon v_1\otimes\dots\otimes v_n\mapsto\frac{1}{n!}\sum_\sigma v_{\sigma(1)}\otimes\dots\otimes v_{\sigma(n)} $$ is a projection onto the space of symmetric $n$-tensors, i.e., $\pi^2=\pi$.

Answer 2

The $\frac1{n!}$ factor doesn't affect the span, but it is customary to include in symmetrification operators, to make it an "average" rather than a "sum". It makes the operation idempotent, along with other nice properties.

Answer 3

See the last page (starting in the 2nd paragraph, but look especially at the final sentence) of http://math.stanford.edu/~conrad/210APage/handouts/tensoralg.pdf or the bottom of page 12 and top of page 13 of http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf.