# Understanding of exterior algebra

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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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
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
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

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."