# Abstract Nonsense

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

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

Exterior Algebra

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Now that we know that $\mathcal{T}(M)$ is a graded $R$-algebra we can start modding out by homogeneous ideals and get new graded $R$-algebras.The one we are going to consider is the ideal $\mathfrak{e}=\left\langle m\otimes m:m\in X\right\}$. This ideal is definitively homogenous and so we can consider the quotient $\mathcal{T}(M) /\mathfrak{e}$ which we denote $\Lambda(M)$ and call the exterior algebra of $M$. We denote the degree $n$ homogeneous part of $\Lambda(M)$ as $\Lambda^n(M)$.

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With rings such that $2\in R^\times$ one can think about $\Lambda(M)$ as being the tensor algebra $\mathcal{T}(M)$ where we have forced “antisymmetry” in the sense that we can now commute simple tensors $v_1\otimes\cdots\otimes v_n$ with any permutation $\sigma\in S_n$ as long as we are willing to concede to multiply the new tensor by $\text{sgn}(\sigma)$.  Indeed, let us denote the coset $v_1\otimes\cdots\otimes v_i+\mathfrak{e}$ as $v_1\wedge\cdots\wedge v_i$,  we’ll call these simple wedges, then the equality

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$(x+y)\otimes(x+y)-x\otimes x-y\otimes y=x\otimes y+y\otimes x$

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and noting that $m\wedge m=0$ for all $m\in M$ we see that

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$0=x\wedge y+y\wedge x$

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so that

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$x\wedge y=-(y\wedge x)$

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We see then that we can change $v_1\wedge\cdots\wedge v_i$ into $v_{\sigma(1)}\wedge\cdots\wedge v_{\sigma(i)}$ with so many transpositions, at each swap picking up a factor of $-1$, and thus after all the total swaps picking up $(-1)^{\#\text{of transpositions of }\sigma}=\text{sgn}(\sigma)$.

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So, why the caveat that $2\in R^\times$? Well, in general I think it should be easy to see that really what modding out by $\mathfrak{e}$ does is give us a new $\mathcal{T}(M)$ where we have forced $v_1\otimes\cdots\otimes v_i\otimes v_{i+1}\otimes\cdots\otimes v_m$ to be zero if $v_i=v_{i+1}$–in other  words, adjacent repetitions are zero. Of course, it’s not hard to see that this always implies the antisymmetry relation as mentioned above, but the two are equivalent if and only if we can divide by $2$–i.e. that $2$ is invertible. Really, if and only if $2$ doesn’t annihilate anything nonzero, but saying $2\in R^\times$ is sufficient for our purposes.

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Now, by definition we can think of $\Lambda^k(M)$, called the $k^{\text{th}}$ exterior product of $M$, as being $\mathcal{T}^k(M)/\mathfrak{e}_k$ where $\mathfrak{e}_k$ is the submodule generated by all elements $m_1\otimes\cdots\otimes m_k\in\mathcal{T}^k(M)$ with two adjacent entries being equal (i.e. just $\mathfrak{e}\cap \mathcal{T}^k(M)$–the $k$ degree homogeneous part of $\mathfrak{e}$). That said, it’s easy to see from the anticommutativity relations described above that really $\mathfrak{e}_k$ is just the ideal generated by all of the elements that look simple tensors where any two elements of the tensor (not necessarily adjacent!) are equal.

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What we’d first like to claim is that $\Lambda^k(M)$ has a nice universal property. Indeed, we know that $\mathcal{T}^k(M)$ allows us to trade $k$-linear maps $M^k\to N$ in for linear maps $\mathcal{T}^k(M)\to N$. What we’d like to claim is that $\Lambda^k(M)$ allows us to trade in $k$-linear alternating maps $M^k\to N$ in for linear maps $\Lambda^k(M)\to N$.

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Recall that an alternating map $f:M^k\to N$ is a $k$-linear map such that $f(m_1,\cdots,m_k)=0$ if any $m_i=m_j$. The classic example of an alternating map is the $n$-linear map $(\mathbb{R}^n)^n\to\mathbb{R}$ given by the determinant (where we are obviously identifying an $n$-tuple of $\mathbb{R}^n$ elements with their associated $n\times n$-matrix).

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Now, our claim is that $\Lambda^k(M)$ for $k\geqslant 2$ satisfies the following universal mapping property: given any $R$-module $N$ and an $k$-linear alternating map $f:M^k\to N$ there exists a unique $R$-linear map $\overline{f}:\Lambda^k(M)\to N$ such that $\overline{f}(m_1\wedge\cdots\wedge m_k)=f(m_1,\cdots,m_k)$.

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Indeed, since the set of all simple wedges generates $\Lambda^k(M)$ as an $R$-module it’s clear that any such map is unique, and so we must merely prove existence. To do this we note that since $f$ is $k$-linear we get an $R$-map $\mathcal{T}^k(M)\to N$ and so it suffices to show that kernel of this map contains $\mathfrak{e}_k$. But, this is clear for if $m_i=m_{i+1}=0$ then $m_1\otimes\cdots m_i\otimes m_{i+1}\otimes\cdots\otimes m_k$ maps to $f(m_1,\cdots,m_i,m_{i+1},\cdots,m_k)=0$. Thus, we see the that the map $\mathcal{T}^k(M)\to N$ descends to an $R$-linear map $\Lambda^k(M)=\mathcal{T}^k(M)/\mathfrak{e}_k\to N$ with the property that

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$m_1\wedge\cdots\wedge m_k=m_1\otimes\cdots\otimes m_k+\mathfrak{e}_k\mapsto f(m_1,\cdots,m_k)$

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as desired. We summarize this as follows:

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Theorem: Let $M$ be an $R$-module (for $R$ a commutative unital ring) and $k\geqslant 2$. Then, for every alternating $k$-linear $R$ map $K:M^k\to N$ (where $N$ is some other $R$ module) there exists a unique $R$-linear map $\overline{K}:\Lambda^k(M)\to N$ with $\overline{K}(m_1\wedge\cdots\wedge m_k)=K(m_1,\cdots,m_k)$

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It’s obvious that since this is a universal mapping property that we could have defined $\Lambda^k(M)$ to be any module satisfying this property, since any two such $R$-modules are necessarily canonically isomorphic.

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Note that this also gives us a way to define $R$-linear maps out of exterior products. Indeed, just like the tensor product, while the simple tensors span $\mathcal{T}^k(M)$ we know that they aren’t, in general, linear dependent so that we can’t define an $R$-map on the simple tensors and “extend by linearity”. Indeed, to construct an $R$-map $\mathcal{T}^k(M)\to N$ we first define a $k$-linear $R$-map $M^k\to N$ and then use the universal property of the tensor product to produce an $R$-linear map $\mathcal{T}^k(M)\to N$ that acts on simple tensors “how we want”.  Similarly, to define maps $\Lambda^k(M)\to N$ we cannot merely define the map on the simple wedges and extend by linearity (for the exact same reason we can’t do this for tensor products). Instead we need to construct an alternating $k$-linear $R$-map $M^k\to N$ and then use the universal property to produce an $R$-linear map $\Lambda^k(M)\to N$ with the desired action on simple wedges.

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Note that thinking about exterior products in this way gives us an interesting way to view the elements of $\Lambda^k(M)$. Indeed, note that an element $m_1\wedge\cdots\wedge m_k$ is nonzero in $\Lambda^k(M)$ if and only if there exists some $R$-module $N$ and some alternating map $M^k\to N$ such that $(m_1,\cdots,m_k)$ is not sent to zero (why?). In particular, this gives us an interesting view of what it should mean for $\Lambda^k(M)$ to be the zero module. It should mean that the only alternating maps out of $M^k$ to any module is the zero map.

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For example, we can now prove easily that if $R$ is an integral domain then $\Lambda^2(\text{Frac}(R))=0$. Indeed, suppose that $N$ is any left $R$-module and $f:\text{Frac}(R)\times\text{Frac}(R)\to N$ is alternating. It’s easy to see that $f\left(\frac{a}{b},\frac{c}{d}\right)=\frac{ac}{bd}f(1,1)$ (the numerators come out for free, and the denominators come out by multiplying on the outside by, say, $\frac{d}{d}$ and bringing the numerator into the second coordinate). But, since $f$ was alternating it follows that $f\left(\frac{a}{b},\frac{c}{d}\right)=0$ and since $\frac{a}{b},\frac{c}{d}$ were arbitrary it follows that $f=0$. Thus, there exists non nonzero $2$-alternating maps out of $\text{Frac}(R)\times\text{Frac}(R)$ from where the above tells us that $\Lambda^2(\text{Frac}(R))=0$ as desired. This tells us, for example, that $\Lambda_\mathbb{Z}^2(\mathbb{Q})=0$ and $\Lambda_{\mathbb{C}[x]}^2(\mathbb{C}(x))=0$.

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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.