# Why is $\pi(\sigma)$ a well defined mapping $V^{\otimes n} \to V^{\otimes n}$?

Given an $$R$$-module V, where $$R$$ is a commutative ring (we even assume $$\mathbb{Q}$$ is a subring, though I don't think that's required for my particular question), we wish to view the wedge product $$V^{\wedge n}$$ as a submodule of $$V^{\otimes n}$$ rather than as a quotient. In attempting to accomplish this, we define a mapping $$\pi\colon S_n\to \operatorname{End}_R(V^{\otimes n})$$ by

$$\pi(\sigma)(\sum_{i=1}^k r_iv_{i1}\otimes v_{i2}\cdots \otimes v_{in}) = \sum_{i=1}^kr_i v_{i\sigma^{-1}(1)}\otimes v_{i\sigma^{-1}(2)}\cdots \otimes v_{i\sigma^{-1}(n)}$$.

I fail to see why $$\pi(\sigma)$$ is a well defined mapping $$V^{\otimes n}\to V^{\otimes n}$$. How would I show that? Elements of $$V^{\wedge n}$$ can typically be written as a linear combination of pure tensors in more than one way. How does one show that whichever linear combination one chooses, the value of the RHS is always the same?

• @coiso Aren't there multiple ways of representing an element of $V^{\otimes n}$, given that the elements are cosets? How do we make sure that our particular choice of representative doesn't affect the value of the right hand side? Commented Sep 13, 2023 at 15:57
• @coiso No, that's exactly my question. Commented Sep 13, 2023 at 16:01
• @coiso I fixed numerous things. My question is why it's well defined. You can restrict to elements of the form $v_1\otimes\cdots \otimes v_n$ if you like. Commented Sep 13, 2023 at 16:06
• If you define $V^{\otimes n}$ to be a quotient of the free vector space on $V^n$, you can first define $\pi(\sigma)$ on the free vector space, show the subspace being quotient'd by is stable, hence $\pi(\sigma)$ is defined on the quotient, and has the given effect on simple tensors. In other words, applying $\pi(\sigma)$ to both sides of the relations (characterizing the multilinearity of the $\otimes$ symbol) yields another valid relation. Commented Sep 13, 2023 at 16:07
• @coiso Could you be more explicit? I know of the lemma saying that if $T: V\to W$ is $R$-linear and $S$ is a submodule of $V$, then the mapping $\overline{T}: V/S\to W$ given by $\overline{T}(x+S) = T(x)$ is well defined and linear. I don't see how to use that here though, since I want to define $\pi$ on $S_n$ which certainly isn't a quotient... Commented Sep 13, 2023 at 16:15

Fix $$\sigma \in S_n$$. Define a map $$V^n\to V^{\otimes n}$$ via $$(v_1,\ldots,v_n)\mapsto v_{\sigma^{-1}(1)} \otimes \ldots \otimes v_{\sigma^{-1}(n)}$$.
Now check that this is multilinear and hence induces by the universal property of tensor products a well-defined linear map $$V^{\otimes n}\to V^{\otimes n}$$ which is the map $$\pi(\sigma)$$ that you want.
(If you only know the universal property of tensor products with two factors, use induction to generalize it to $$n$$ factors.)
• Thank you. This, in addition to the lemma I stated in the comments (which, by the way, should include the extra hypothesis $T(S) = \{0\}$) did the trick. Commented Sep 14, 2023 at 4:33
• Now I'm stuck on showing that it's a group homomorphism. I'm thinking it would suffice to show that given permutations $\sigma$ and $\tau$, one has $v_{\sigma^{-1}(\tau^{-1}(1))}\otimes\cdots\otimes v_{(\sigma^{-1}(\tau^{-1}(n))} = v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)}+v_{\tau^{-1}(1)}\otimes\cdots\otimes v_{\tau^{-1}(n)}$. Any ideas? Commented Sep 14, 2023 at 14:06
• @SimonSMN no, you need to show that $\pi(\sigma \tau)=\pi(\sigma) \circ \pi(\tau)$ Commented Sep 14, 2023 at 14:48
• But how is $\operatorname{End}_R(V^{\otimes n})$ then a group w r t composition? Not every element in it has an inverse, does it? Commented Sep 14, 2023 at 15:00