# Abstract Nonsense

## The Tensor Algebra and Exterior Algebra (Pt. IV)

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

Exterior Products of Free Modules

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We’d now like to discuss how everything works out when we are talking about exterior products of free modules.

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In particular, we know that if $M$ is free, then by virtute of the fact that the tensor product commutes with coproducts we know that $\mathcal{T}^k(M)$ is free. But, it is rookie-mistake number one to think that quotients of free modules (even by free submodules) are necessarily free (is $\mathbb{Z}/2\mathbb{Z}$ free?!) and so there is no a priori reason to think that freeness of $M$ should imply the freeness of $\Lambda^k(M)$.

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Now, while not explicitly stated, it should be fairly obvious that if $\{v_1,\cdots,v_m\}$ is a spanning set for $M$ then $\{v_{i_1}\wedge\cdots\wedge v_{i_k}\}_{i_1,\cdots,i_k\in[m]}$ is a spanning set for $\Lambda^k(M)$ since that set, with wedges replaced with tensors, is a spanning set for $\mathcal{T}^k(M)$ of which $\Lambda^k(M)$ is a quotient. Of course, there is a huge amount of redundancies, in fact, most of these wedges are zero. For example, we can immediately replace this set with $\{v_{i_1}\wedge\cdots\wedge v_{i_k}\}$ where now the indices still live in $[m]$ but now we require that $i_\ell\ne i_j$ for any $\ell,j$. But, once again this is still redundant for we have indexed by injections $[k]\to[m]$, but by virtue of anticommutativity, we can index by combinations of $k$ elements of $m$. In particular, $\{v_{i_1}\wedge\cdots\wedge v_{i_k}\}$ spans $\Lambda^k(M)$ where $i_1<\cdots. In particular, we see that if $M$ is spanned by $m$ elements then $\Lambda^k(M)$ is spanned by $\displaystyle {m\choose k}$ elements. In particular, we see the following:

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Theorem: If $M$ is spanned by $m$ elements and $k>m$ then $\Lambda^k(M)=0$

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This makes sense with our above interpretation of $\Lambda^k(M)$ being zero. Indeed, this should say that there are no nonzero alternating $k$-linear $R$-maps out of $M^k$. But, it of course suffices to check that if $f$ is such a map that $f(v_{i_1},\cdots,v_{i_k})=0$ for the $v_{i_j}$ in the spanning set of $M$. But , our old pal The Pigeonhole Principle (easily one of the most under appreciated theorems in mathematics!) tells us that since $k that at least two of these $v_{i_j}$‘s must be equal and so $f(v_{i_1},\cdots,v_{i_k})=0$.

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What we’d like to show is that if $M$ is indeed free, so that we can take this set of $v_i$‘s to be a basis, then this generating set is also a basis. The key to this is that when $M$ is free we shall be able to embed $\Lambda^k(M)$ back into $\mathcal{T}^k(M)$–a strange concept, embedding a quotient back into the space. The embedding should look familiar either if you are familiar with the more classic approach to multilinear algebra (via the alternating map) or if you know Leibniz expansion of the determinant. Namely, we are going to define an embedding $\Lambda^k(M)\hookrightarrow\mathcal{T}^k(M)$ in such a way that

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$\displaystyle v_1\wedge\cdots \wedge v_k\mapsto \sum_{\sigma\in S_k}\text{sgn}(\sigma)(v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(k)})\quad\mathbf{(1)}$

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To see that such a map even exists we note that the mapping $M^k\to\mathcal{T}^k(M)$ defined as above (except we are mapping $(m_1,\cdots,m_k)$ to this sum) is alternating. Indeed, it’s easy to see that this map is multilinear (being a sum of multilinear maps) and is alternating for

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$\displaystyle v_{\tau(1)}\wedge\cdots\wedge v_{\tau(k)}\mapsto \sum_{\sigma\in S_k}\text{sgn}(\sigma)(v_{(\sigma\circ\tau)(1)}\otimes\cdots\otimes v_{(\sigma\circ\tau)(k)})$

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Now, we write $\text{sgn}(\sigma)=\text{sgn}(\tau)^{-1}\text{sgn}(\sigma\circ\tau)$ (recalling that $\text{sgn}$ is a group map $S_k\to\{-1,1\}$) and since $\text{sgn}(\tau)$ is idempotent this is just $\text{sgn}(\sigma)=\text{sgn}(\tau)\text{sgn}(\sigma\circ\tau)$. So, we see that $(m_{\tau(1)},\cdots,\cdots,m_{\tau(k)})$ is mapped to

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$\displaystyle \text{sgn}(\tau)\sum_{\sigma\in S_k}\text{sgn}(\sigma)(v_{\sigma(1)}\wedge\cdots\wedge v_{\sigma(k)})$

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Thus, we are afforded an $R$-linear map $\Lambda^k(M)\to \mathcal{T}^k(M)$ with $\mathbf{(1)}$ holding. What we’d like to now show is that if $M$ is free this is actually an embedding. Indeed, let $e_1,\cdots,e_m$ be a basis for $M$. Let $\displaystyle \sum_{I}\alpha_I e_I\in\Lambda^k(M)$ where $I$ runs over indices $1\leqslant i_1<\cdots<_k\leqslant m$ and $e_I$ evidently means $e_{i_1}\wedge\cdots\wedge e_{i_k}$ if $I=(i_1,\cdots,i_k)$. Now, if this goes to zero we have the relation

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$\displaystyle \sum_{\sigma\in S_k}\sum_{I}\text{sgn}(\sigma)\alpha_I e_{\sigma(I)}=0$

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That said, from the basic theory of tensor products we know that $\{e_{\sigma(I)}\}$, as $\sigma$ and $I$ run over their respective indices, form a basis for the free $R$-module $\mathcal{T}^k(M)$ which tells us that $\text{sgn}(\sigma)\alpha_I=0$ for each $I$ and each $\sigma$–but this in particular tells us that $\alpha_I=0$ for all $I$. Thus, we see that the kernel of our map is trivial and thus we have an honest to god embedding. Now, note that the image of the vectors $\{e_I\}$ are linearly independent (for if $I$ and another index $J$ aren’t equal we can see that the image of $e_I$ and $e_J$ will contain different basis vectors!) and thus since we are dealing with an embedding this tells us that that $\{e_I\}$ are linearly independent in $\Lambda^k(M)$. Thus, we may conclude:

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Theorem: Let $M$ be a free $R$-module with basis $\{e_1,\cdots,e_n\}$. Then, $\Lambda^k(M)$ is a free $R$-module with basis $\{e_{i_1}\wedge\cdots\wedge e_{i_k}\}$ as the indices are chosen arbitrarily according to the condition $1\leqslant i_1<\cdots. In particular, we see that $\text{rank }\Lambda^k(M)$ is $\displaystyle {n\choose k}$.

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This is actually enables us to evaluate the full exterior algebra, as a module. Indeed:

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Theorem: Let $M$ be a f.g. free $R$-module of rank $n$. Then, $\Lambda(M)$ is a free $R$-module of rank $2^n$.

Proof: We know that $\displaystyle \Lambda(M)$ is $\displaystyle \bigoplus_{k=0}^{\infty}\Lambda^k(M)$. Now, each $\Lambda^k(M)$ is free of rank $\displaystyle {n\choose k}$ and thus $\Lambda(M)$ is free of rank

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$\displaystyle \sum_{k=0}^{\infty}{n\choose k}=\sum_{k=0}^{n}{n\choose k}=2^n$

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$\blacksquare$

While this embedding seemed incidental it shall come back to important for us when we finally get to the differential forms stuff that we are working towards.

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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.