# Abstract Nonsense

## Some Natural Identifications

Point of Post: In this post we take our discussion last time concerning tensor algebras and their quotients to the case of vector spaces, and show that there are simple identifications that can be made. We also discuss the notion of orientations and volume forms.

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Motivation

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We now are able to take the very formal, very general way of defining tensor powers/exterior powers and show how they relate to much tamer objects–multilinear and alternating forms. The main reason for doing this is, as mentioned before, to help explain the somewhat confusing notation used in some differential geometry textbooks. For example, they merely denote the space of alternating $k$-forms on $V$ as $\Lambda^k(V^\ast)$. But, why? This notation makes no sense, especially considering that this notation has other meaning in mathematics (the $k^{\text{th}}$ exterior power of the dual space of $V$). They should really only conflate these two if there is some sort of natural isomorphism between these two. Well, to state the obvious, they are naturally isomorphic. But, this is never stated, let alone proven, in differential geometry textbooks. This is the point of this post.

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Natural Identification of $\mathcal{T}^k(V^\ast)$ with $\text{Mult}_k(V)$

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We begin with some finite dimensional vector space $V$ over a field $F$. We recall that $V\cong V^\ast$ where $V^\ast$, the dual space, is $\text{Hom}_F(V,F)$. Moreover, we recall that if $v_1,\cdots,v_n$ is a basis for $V$ then $v_1^\ast,\cdots,v_n^\ast\in V^\ast$ defined by $v_i^\ast(v_j)=\delta_{i,j}$ is the dual basis–obviously a basis for $V^\ast$.

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Recall also that $\text{Mult}_k(V)$ is the space of all $k$-linear forms $V^k\to F$–if $k=0$ we just take the forms to be $F$. Our first claim is that there is a natural identification $\text{Mult}_k(V)\leftrightarrow \mathcal{T}^k(V^\ast)$.  Indeed, define $(V^\ast)^k\to\text{Mult}_k(V)$ by saying that $(\varphi_1,\cdots,\varphi_k)$ is sent to a funtion $V^k\to F$ which acts as

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$(v_1,\cdots,v_k)\mapsto \varphi_1(v_1)\cdots\varphi_k(v_k)$

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It’s evident not only that this function associated to $(\varphi_1,\cdots,\varphi_k)$ is a $k$-linear form but also that the association itself is $k$-linear. Thus, we get a map $\Phi_V:\mathcal{T}^k(V^\ast)\to\text{Mult}_k(V)$ where

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$\Phi_V(\varphi_1\otimes\cdots\otimes\varphi_k)(v_1,\cdots,v_k)=\varphi_1(v_1)\cdots\varphi_k(v_k)$

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It’s easy to see that this is an $F$-map. To prove that this is an injection suppose that $\Phi_V(\varphi_1\otimes\cdots\otimes \varphi_k)$ is the zero map $V^k\to F$.

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Now, suppose that $W$ is another $F$-space and $T:V\to W$ is a given map. We know that we get a map $T^\ast:W^\ast\to V^\ast$ by precomposing any functional $W\to F$ with $T$. But, we then get from the construction of the tensor product algebra a map $\mathcal{T}^k(T^\ast):\mathcal{T}^k(W^\ast)\to\mathcal{T}^k(V^\ast)$ given on simple tensors by

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$\mathcal{T}^k(T^\ast)(\varphi_1\otimes\cdots\otimes \varphi_k)=T^\ast(\varphi_1)\otimes\cdots\otimes T^\ast(\varphi_k)=(\varphi_1\circ T)\otimes\cdots\otimes (\varphi_k\circ T)$

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Similarly, we get a map $\mathcal{M}(T):\text{Mult}_k(W)\to\text{Mult}_k(V)$ defined by $\mathcal{M}(T)(K)(v_1,\cdots,v_k)=K(T(v_1),\cdots,T(v_k))$ for every $k$-linear form $K$. Note then that for each simple tensor $\varphi_1\otimes\cdots\otimes \varphi_k$ and each vector $(v_1,\cdots,v_k)\in V^k$ we have that

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\begin{aligned}\left[(\Phi_V\circ\mathcal{T}^k(T^\ast))(\varphi_1\otimes\cdots\otimes \varphi_k)\right](v_1,\cdots,v_k) &=(\Phi((\varphi_1)\circ T)\otimes\cdots\otimes (\varphi_k\circ T))(v_1,\cdots,v_k)\\ &=\varphi_1(T(v_1))\cdots\varphi_k(T(v_k))\end{aligned}

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and

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\begin{aligned}\left[(\mathcal{M}(T)\circ\Phi_W)(\varphi_1\otimes\varphi_k)\right](v_1,\cdots,v_k) &=(\Phi_W(v_1\otimes\cdots\otimes v_k))(T(v_1),\cdots,T(v_k))\\ &=\varphi_1(T(v_1))\cdots\varphi_k(T(v_k)\end{aligned}

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So that the diagram

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$\begin{array}{ccc}\mathcal{T}^k(W^\ast) & \overset{\mathcal{T}^k(T^\ast)}{\longrightarrow} & \mathcal{T}^k(V^\ast)\\ ^{\Phi_W}\big\downarrow & & \big\downarrow^{\Phi_V}\\ \text{Mult}_k(W) & \underset{\mathcal{M}(T)}{\longrightarrow} & \text{Mult}_k(V)\end{array}$

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Commutes, and thus we may finally conclude the following:

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Theorem: Let $V$ be a finite dimensional vector space. Then, the map $\Phi:\mathcal{T}^k(V^\ast)\to\text{Mult}_k(V)$ as defined above is a natural isomorphism.

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Similarly, we note that there is a natural isomorphism $\Psi_V:\mathcal{T}(V^\ast)\to\text{Mult}(V)$ where $\text{Mult}(V)$ is $\displaystyle \bigoplus_{j=0}^{\infty}\text{Mult}_k(V)$. This is an isomorphism of $F$-algebras when we define the multiplication, auspiciously denoted $\otimes$, in $\text{Mult}(V)$ by saying that if $f\in\text{Mult}_i(V)$ and $g\in\text{Mult}_j(V)$ then $f\otimes g\in\text{Mult}_{i+j}(V)$ is given by

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$(f\otimes g)(v_1,\cdots,v_i,w_1,\cdots,w_j)=f(v_1,\cdots,v_i)g(w_1,\cdots,w_j)$

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This is easy to prove.

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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.