# Abstract Nonsense

## The Tensor Algebra and the Exterior Algebra (Pt. I)

Point of Post: In this post we introduce the basic notions of the tensor algebra, graded algebras/rings, and the exterior algebra.

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Motivation

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If there is anyone (and I do mean anyone) who reads my blog regularly, they should know that I have started blogging about complex analysis. It was going fine and dandy until–blam!–I hit Cauchy’s Theorem. Why was this a problem? Well, I was going to discuss how Cauchy’s theorem is simple for holomorphic functions (as I have defined them, as being $C^1$) since we could apply Green’s theorem. But, then it hit me, I didn’t really remember how the proof of Green’s/Stokes theorem went. Thus, I decided that I should review all of the basic differential forms/Stokes theorem stuff and post it on my blog. In preparation for this I started looking at books, upon books to brush up on this stuff. One of the things that I found most confusing was the lack of algebraic rigour concerning the machinery involved. In particular, books either do one of two things. They either don’t talk about the more conceptually difficult algebra they are using, which while easier to understand (to someone not algebraically inclined!) makes certain constructions/operations seem forced, unnatural. Or, they use the higher level stuff and make identifications with the lower level stuff without explaining properly what this identification is, and why it makes sense.

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Thus, before I actually start talking about differential forms I want to discuss the algebra that is going to be needed. This post is devoted to the higher level stuff–the tensor algebra and exterior powers. The tensor algebra is a very natural constructions since it is the “free algebra” over a given module–it’s just what you get when you just multiply vectors in a module, without any restrictions. One can think about the tensor product as being noncommutative polynomials where the indeterminates are the elements of the module you started with. The exterior algebra can be thought about algebraically as the freest algebra on a module where every indeterminate squares to zero. While this is slightly less interesting than the tensor algebra, its smaller pieces (the exterior powers of the module) is what is really interesting for they are to alternating multilinear maps what the tensor product is to just plain old multilinear maps. In other words, the exterior powers will be the modules that let us trade in alternating multiliner maps for linear ones in the “freest” possible way.

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I will be following heavily from [6], making altercations and additions as I see fit.

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The Tensor Algebra

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Let $R$ be a commutative unital ring and $M$ a left $R$-module. Let $\mathcal{T}^j(M)$ denote the $j$-fold self tensor product $M^{\otimes j}$ (with $\mathcal{T}^0(M)=R$) and define the tensor algebra $\mathcal{T}(M)$ to be

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$\displaystyle \mathcal{T}(M)=\bigoplus_{j=0}^{\infty}\mathcal{T}^j(M)$

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We will not think of the elements of $\mathcal{T}(M)$ as tuples but as finite formal sums of elements of the $\mathcal{T}^j(M)$s. Clearly this is an $R$-module, but to deserve its namesake it better have some kind of multiplication. To define such a multiplication it suffices to define a multiplication on a generating set. Namely, define the multiplication, which we shall denote $\otimes$, on $m_1\otimes\cdots\otimes m_i$ and $m'_1\otimes\cdots\otimes m'_j$ by

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$(m_1\otimes\cdots\otimes m_i)\otimes(m'_1\otimes\cdots\otimes m'_j)=m_1\otimes m_i\otimes m'_1\otimes\cdots\otimes m'_j$

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I will leave it to you to show that if we extend this multiplication by distributivity to all of $\mathcal{T}(M)$ that we do get a well-defined ring structure on $\mathcal{T}(M)$ that satisfies

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$\mathcal{T}^i(M)\otimes\mathcal{T}^j(M)\subseteq\mathcal{T}^{i+j}(M)$

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(where on the left this denotes the set of all pairwise multiplications). Moreover, it’s clear that for any $x,y\in\mathcal{T}(M)$ and $r\in R$ one has that

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$r(x\otimes y)=(rx)\otimes y=x\otimes (ry)$

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so that $\mathcal{T}(M)$ really is an, honest to god, $R$-algebra.

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Given two $R$-modules $M$ and $N$ and an $R$-map $f:M\to N$ we get an $R$-algebra map $\mathcal{T}(f):\mathcal{T}(M)\to \mathcal{T}(N)$ given on simple tensors by $\mathcal{T}(f)(m_1\otimes\cdots\otimes m_j)=f(m_1)\otimes\cdots\otimes f(m_j)$. I’ll let you verify that this really is an $R$-algebra map. It’s easy to see that $\mathcal{T}$ (on maps) preserves compositions and identity and, moreover, preserves the sum of linear maps. Thus, we may conclude that $\mathcal{T}:R\text{-}\mathbf{Mod}\to R\text{-}\mathbf{Mod}$ is an additive functor.

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The reason why this algebra is so fundamental, such a natural construction is as follows. Suppose that $A$ is an $R$-algebra and we have an $R$-map $f:M\to A$. We note then that the map $M^j\to A$ given by $(m_1,\cdots,m_j)\mapsto f(m_1)\cdots f(m_j)$ is $R$-bilinear and so extends to an $R$-map $F_j:\mathcal{T}^j(M)\to A$. By the definition of direct sum (being a coproduct) this then gives us an $R$-map $F:\mathcal{T}(M)\to A$ defined on simple tensors as

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$F(m_1\otimes\cdots\otimes m_j)=F_j(m_1\otimes\cdots\otimes m_j)=f(m_1)\cdots f(m_j)$

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It’s easy to see that this really is a ring map, and thus an $R$-algebra map. Moreover, it’s clear that such map on simple tensors is unique. Thus, we realize that there is a bijection

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$\text{Hom}_R(M,A)\cong \text{Hom}_{R\text{-}\mathbf{Alg}}(\mathcal{T}(M),A)$

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This suggests that there may be some sort of adjunction hiding in the background. And, indeed, this is true. Namely, the functor $\mathcal{T}:R\text{-}\mathbf{Mod}\to R\text{-}\mathbf{Alg}$ is a left adjoint to the forgetful functor $U:R\text{-}\mathbf{Alg}\to R\text{-}\mathbf{Mod}$. This is just a bunch of diagram verifications, which I leave to you.

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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.

May 10, 2012 -

1. […] The Tensor Algebra and the Exterior Algebra (Pt. II) Point of Post: This is a continuation of this post. […]

Pingback by The Tensor Algebra and the Exterior Algebra (Pt. II) « Abstract Nonsense | May 10, 2012 | Reply

2. Thank you so much for doing this. I’ve taken the second quarter of graduate algebra at my university, and in the course we defined the wedge product in terms of tensor product (in the context of arbitrary modules, of course), and then this quarter, as I’m also in the third quarter of undergraduate real analysis following Baby Rudin, I’m seeing wedge again in the context of differential forms, and Rudin completely left out the algebraic meaning behind everything. It’s been incredibly unsatisfying reading Rudin as a result.

Comment by Reeve | May 10, 2012 | Reply

• You are more than welcome! I find it very annoying that basic graduate texts seem to go to lengths to homogenize the subjects. I mean, I can see reading Rudin and going “Yes, I am doing analysis, but what is this algebra we are skirting by?”–it gets frusturating.

Stay tuned for what should be, hopefully, a nice supplement to what you are currently learning!

I wish you the best of luck! Feel free to ask me any questions if they should so arise.

Best,
Alex

Comment by Alex Youcis | May 10, 2012 | Reply