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Consider the following definition from Loring W. Tu's An Introduction to Manifolds:

For a finite-dimensional vector space $V$, say of dimension $n$, define $$A_*(V)=\oplus_{k=0}^{\infty}A_k(V)=\oplus_{k=0}^{n}A_k(V)$$ where $A_0(V)={\mathbb R}$, and $A_k(k>0)$ denotes the set of all alternating $k$-linear functions $f$ on $V$, i.e., $$f:V^k\to{\mathbb R},\qquad f(v_{\sigma(1),\cdots,v_{\sigma(k)}})=(\text{sgn}\sigma)f(v_1,\cdots,v_k) \quad\text{for all} \quad\sigma\in S_k.$$ With the wedge product of multicovectors as multiplication, $A_*(V)$ becomes an anticommutative graded algebra, called the exterior algebra or the Grassmann algebra of multicovectors on the vector space $V$.

By definition of graded algebra, $A_*(V)$ has the structure of a vector space. But I don't understand what does the element of $A_*(V)$ look like. For example, if $f\in A_2(V)$ and $g\in A_3(V)$, then what is $f+g$? Since domains of $f$ and $g$ are of different dimensions, how can one "add" them?

So here is my question:

What does the element of $A_*(V)$ look like? And what's the addition of the vector space?


According to the comments, the question above is actually a matter of understanding of the "direct sum". In stead of putting another post, I would like to ask a closed related question here:

How many different definitions are there of exterior algebra and how are they equivalent to each other?

I totally don't understand the one I learn from wikipedia. Any recommendation of a complete treatment of the subject?

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7  
The addition is formal addition; that's what the direct sum means. Are you familiar with direct sums? –  Qiaochu Yuan Jul 7 '11 at 21:02
    
@Qiaochu: +1. Hmm, I thought it the wrong way. I should have thought about the Cartesian product of the underlying sets. Then instead of writing $f+g$ in $A_*$ in the case of my example, it is supposed to be $(0,f,0,\cdots,0)+(0,0,g,0,\cdots,0)$. Correct? –  Jack Jul 7 '11 at 21:10
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More or less, but the Cartesian product is not quite what you want for an infinite direct sum: rather you want the subspace of the Cartesian product where all but finitely many components are zero. –  Qiaochu Yuan Jul 7 '11 at 21:13
    
@Qiaochu: Hmm, just as the addition in $l_p$. Btw, I appreciate your previous link for the lecture note, you deleted it, though.:-) –  Jack Jul 7 '11 at 21:21
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Well, here the direct sum and the Cartesian product really are the same, because all summands $A_k(V)$ with $k > n$ are zero. Anyway, this makes me think of a "joke" I read on some blog: it's false that you can't add apples and oranges. You can add them in the free abelian group on an apple and an orange. It's not hilarious, but it's insightful: the point is that the sum of two algebra elements of different dimensions is a somewhat strange quantity, like the sum of an apple and an orange. Often in graded algebras -- particularly here -- one is most interested in homogeneous elements. –  Pete L. Clark Jul 7 '11 at 21:45
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1 Answer

up vote 2 down vote accepted

There are two views on graded rings, the view taken by algebraists and the view taken by topologists. To an algebraist, an element of the graded ring is just a formal linear combination of elements of each degree (we say an element of a single degree is homogenous). To a topolgist, you never actually add elements of different degrees. You can multiply two things of different degrees, but adding them gives nonsense. Not nonsense in that it can't be interpreted algebraically, but nonsense in the sense that your homogenous elements of your ring correspond to something not purely algebraic while non-homogenous elements don't.

This is a perfect example of why topologists take the approach they do. Given two alternating $n$-multilinear functions $f$ and $g$, we can define $f+g$ to be another alternating $n$-multilinear function. However, there is no good way to make sense of the "sum" of two functions that take in a different number of elements. So the solution, as unhelpful as it might sound is "don't add elements of different degrees, or if you do, don't try to give them meaning."


For your second question, I believe that Lee's introduction to smooth manifolds has a treatment of the exterior algebra in roughly the way that wikipedia does, but likely in enough detail that you can better understand it. The basic idea is that, instead of working with functions (where the multiplication ends up being funny), we just formally define a multiplication of vectors so that we can write "words" in our vectors, and we throw in just the relations that would make the product skew symmetric, and we let that be the alternating algebra. Understanding the construction requires understanding tensor products, which can be tricky the first time you see them.

You can find a quick definition and proof of the basic properties of the exterior algebra here, but I do not know that it will be much more helpful than the wikipedia article.

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