# Abstract Nonsense

## Some Natural Identifications (Pt. II)

Point of Post: This is a continuation of this post.

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Natural Identification of $\Lambda^k(V^\ast)$ with $\text{Alt}_k(V)$

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We now show how to naturally identify $\Lambda^k(V^\ast)$ with the space $\text{Alt}_k(V)$ of alternating $k$-forms $V^k\to F$. In particular, we define the map $P{V,k}:\mathcal{T}^k(V)\to \text{Alt}_k(V)$ as follows:

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$\displaystyle \varphi_1\otimes\cdots\otimes \varphi_k\mapsto \sum_{\sigma\in S_k}\text{sgn}(\sigma)\Psi_V(\varphi_{\sigma(1)}\otimes\cdots\otimes \varphi_{\sigma(k))}$

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When $V$ is clear we shall suppress the $V$ and merely write $P_k$. This map is well-defined (in the sense that we can extend this map on simple tensors to a map on all of $\mathcal{T}^k(V)$) since it’s the composition of the map $\Psi_{V,k}$ with a map $\text{Mult}_k(V)\to\text{Alt}_k(V)$. Note that this map really is linear, and moreover that the image of the map does, in fact, like in $\text{Alt}_k(V)$. We really only need to explain this second point, and we really only need to check this on the image of the simple tensors. To check this we let $\tau\in S_k$ and calculate as follows:

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\displaystyle \begin{aligned}P_k(\varphi_1\otimes\cdots\otimes \varphi_k)(v_{\tau(1)},\cdots,v_{\tau(k)}) &= \sum_{\sigma\in S_k}\text{sgn}(\sigma)\varphi_{\sigma(1)}(v_{\tau(1)})\cdots\varphi_{\sigma(k)}(v_{\tau(k)})\\ &= \sum_{\sigma\in S_k}\text{sgn}(\sigma\circ\tau)\varphi_{\sigma(\tau(1))}(v_{\tau(1)})\cdots\varphi_{\sigma(\tau(k))}(v_{\tau(k)})\\ &=\text{sgn}(\tau)\sum_{\sigma\in S_k}\text{sgn}(\sigma)\varphi_{\sigma(\tau(1))}(v_{\tau(1)})\cdots\varphi_{\sigma(\tau(k))})(v_{\tau(k)})\\ &=\text{sgn}(\tau)\sum_{\sigma\in S_k}\text{sgn}(\sigma)\varphi_{\sigma(1)}(v_1)\cdots\varphi_{\sigma(k)}(v_k)\\ &=\text{sgn}(\tau)P_k(\varphi_1\otimes\cdots\otimes \varphi_k)(v_1,\cdots,v_k) \end{aligned}

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The key being that, using Cayley’s theorem, we know that for any function $f:S_k\to F$ one has that $\displaystyle \sum_{\sigma\in S_k}f(\sigma)=\sum_{\sigma\in S_k}f(\sigma\circ\tau)$.

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What we now claim is that $\ker P_k=\mathfrak{e}_k$ so that $P_k$ descends to an isomorphism $\Lambda^k(V^\ast)=\mathcal{T}^k(V^\ast)/\mathfrak{e}_k\xrightarrow{\approx}\text{Alt}_k(V)$.

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To see that $\mathfrak{e}_k\subseteq\ker P_k$ suppose that $\varphi_1\otimes\cdots\otimes\varphi_k$ is such that $\varphi_i=\varphi_j$. Now, since $[S_k:A_k]=2$ and $(i,j)\notin A_k$ we have that $S_k=A_k\sqcup (i,j)A_k$. Thus, we can write

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\displaystyle \begin{aligned}\sum_{\sigma\in S_k}\text{sgn}(\sigma)\Psi_{V,k}(\varphi_{\sigma(1)}\otimes\cdots\varphi_{\sigma(k)}) &= \sum_{\sigma\in A_k}(\text{sgn}(\sigma)+\text{sgn}((i,j)\sigma))\Psi_{V,k}(\varphi_{\sigma(1)}\otimes\cdots\varphi_{\sigma(k)})\\ &=0 \end{aligned}

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where the last equality follows since

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$\text{sgn}((i,j)\sigma)=\text{sgn}((i,j))\text{sgn}(\sigma)=-\text{sgn}(\sigma)$

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The reverse inclusion is slightly more difficult, but is left to you to figure out (just plug in the appropriate values to show that some two of them must be equal).

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Now, as was stated before we have that $P_k$ descends to an isomorphism $\Lambda^k(V^\ast)\to\text{Alt}_k(V)$ which we also denote as $P_k$.

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Now, I leave it to anyone interested that $P_k$ is natural–this is exceedingly easy since $\Psi_{V,k}$ is natural. Thus, with this we have the following:

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Theorem: Let $V$ be a finite dimensional vector space, then $P_{V,k}:\Lambda^k(V^\ast)\to\text{Alt}_k(V)$ is a natural isomorphism.

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Now, we see that this isomorphism allows us define a product, the wedge product $\omega$,  on the space of all alternating multilinear forms $\displaystyle \text{Alt}(V)=\bigoplus_{k\geqslant0}\text{Alt}_k(V)$ defined by $\displaystyle \omega\wedge\eta=\frac{1}{k!r!}P_{k+r}(P_k^{-1}(\omega)\wedge P_r^{-1}(\eta))$. Explicitly, this comes out to the definition that

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$\displaystyle (\omega\wedge\eta)(v_1,\ldots,v_k,v_{k+1},\ldots,v_{r+k})=\frac{1}{r!k!}\sum_{\sigma\in S_{r+k}}\text{sgn}(\sigma)\omega(v_{\sigma(1)},\ldots,v_{\omega(k)})\eta(v_{\sigma(k+1)},\ldots,v_{\sigma(k+r)})$

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Since this product is defined so that $P_V:\Lambda(V^\ast)\to\text{Alt}(V)$ is an isomorphism of $\mathbb{R}$-algebras we see if we start with an $\mathbb{R}$-algebra generating set for $\Lambda(V^\ast)$ then its image under $P_V$ is an $\mathbb{R}$-algebra generating set for $\text{Alt}(V)$.

Hom, Tensor, and Dual space Duality and the Trace Map

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Unsurprisingly this section of this post shall be based around a natural isomorphism involving $\text{Hom},\otimes$, and $^\ast$ and how this can be used to give a coordinate free definition of the trace map. Throughout the following $V$ and $F$ shall have the same meaning as in the above section, $W$ will be another finite dimensional $F$-space, and $\text{Hom}_F(V,W)$ as well as $V\otimes_FW$ shall be denoted $\text{Hom}(V,W)$ and $V\otimes W$ respectively. Lastly, for purely pedagogical reasons we denote $\varphi(v)$ as $\langle\varphi,v\rangle$ for $\varphi\in V^\ast$ and $v\in V$–it is important to note that $\langle-,-\rangle:V^\ast\times V\to F$ is bilinear.

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The isomorphism we would like to prove is $\text{Hom}(V^\ast,W)\cong V\otimes W$. From a purely dimension counting, variance checking, point of view this makes sense. Indeed, $\text{Hom}(V,W)\cong V\otimes W$ purely from a dimension check–they both have dimension $\dim(V)\dim(W)$. That said, they clearly can’t be naturally isomorphic for, if they were, $\text{Hom}(\bullet,W)$ and $\bullet\otimes W$ would have to be isomorphic functors, but this can’t be true since the first is contravariant and the second is covariant. For those who aren’t so comfortable with categorical language this really just means that if $V\to V'$ is a linear map we get linear maps $\text{Hom}(V',W)\to\text{Hom}(V,W)$ and $V\otimes W\to V'\otimes W$–the fact that they are going in opposite directions (one keeps $V$ in the domain and the other puts it in the codomain) tells us there can’t be a natural isomorphism. So, how do we fix this? The obvious way to fix this would be to compose $\text{Hom}(\bullet,W)$ with some kind of contravariant functor…well, this isn’t guaranteed to fix it but it is definitely necessary. Moreover, we need the functor we compose with to be dimension preserving. Well, anyone who has done linear algebra long enough should know the classic example of this: the dual space functor. Thus, we make a guess that the functor $\bullet\otimes W$ should be isomorphic to the functor with $V\mapsto \text{Hom}(V^\ast,W)$.

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While this really doesn’t make obvious the fact that $\text{Hom}(V^\ast,W)\cong V\otimes W$ it at least gives a plausibility argument (although a bad one). Regardless, let’s actually get down to defining this isomorphism.

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

[5] Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.