Embedding from exterior product to tensor product space

I came across the following:

For a basis $\phi_1,\dots,\phi_n$ of $V$ there is a natural embedding $V^{\wedge n}\hookrightarrow V^{\otimes n}$ defined as $$(\phi_1\wedge \cdots \wedge \phi_n) \mapsto\frac{1}{\sqrt{n!}}\sum\limits_{\sigma \in \mathcal{S}_n} \operatorname{sgn}(\sigma) (\phi_{\sigma(1)}\otimes \cdots\otimes \phi_{\sigma(n)})$$

Why is there a square root? In Spivak's book he defines the map $V^{\otimes n}\to V^{\wedge n}$ $$ \operatorname{Alt}(\phi_1\otimes \cdots \otimes \phi_n) =\frac{1}{n!}\sum\limits_{\sigma \in \mathcal{S}_n} \operatorname{sgn}(\sigma) (\phi_{\sigma(1)}\otimes \cdots\otimes \phi_{\sigma(n)}) $$

Solution 1:

It is true very generally (say $V$ is a free module over an arbitrary commutative ring) that the map $$ i_n: \Lambda^n V \to V^{\otimes n}$$ defined by $$ i_n(v_1 \wedge \cdots \wedge v_n) = \sum_{\sigma \in S_n} (-1)^\sigma v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(n)}$$ is an embedding: in fact it is an isomorphism onto the space of skew-symmetric tensors. There is a natural map in the other direction defined by $$ \pi_n: V^{\otimes n} \to \Lambda^n V, \quad \pi_n(v_1 \otimes \cdots \otimes v_n) = v_1 \wedge \cdots \wedge v_n,$$ and it is easy to see that $\pi_n \circ i_n$ is multiplication by $n!$ on $\Lambda^n V$. Therefore as long as the base ring has characteristic zero (say we are working over $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$) then both $\pi_n$ and $i_n$ are isomorphisms (again, when considered as maps to and from the subspace of skew-symmetric tensors in $V^{\otimes n}$).

If the author is happy to work over fields of characteristic zero, they may want to have $\pi_n \circ i_n$ compose to the identity instead of $n!$, therefore one of those maps needs to be divided by $n!$, or both divided by $\sqrt{n!}$ for symmetry.