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In Tu's An Introduction to Manifolds; an exterior algebra is defined as follows:

For a finite dimensional vector space $V$, say of dimension $n$, define $$A_*(V) = \bigoplus_{k=0}^\infty A_k(V) = \bigoplus_{k=0}^n A_k(V)$$

where $A_k(V)$ is the space of alternating $k$-covectors of the space $V$. Why can we truncate the sum from $\infty$ to $n?$ Wouldn't this only be true if $l$-covectors where $l > n$ were somehow expressible as sums of $m$-covectors for $m \leq k$? Is this true?

QCD_IS_GOOD
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  • @user7530 Tu defines $k$-covectors as alternating $k$-linear maps, so they are alternating by definition according to Tu - ill update question to avoid confusion – QCD_IS_GOOD Apr 14 '17 at 04:04

2 Answers2

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$V$ has dimension $n$, and $A_k(V)$ has dimension $\binom{n}{k}$. In particular, when $k > n$, $A_k(V)$ is zero-dimensional. (This is Corollary 3.31 in my version of the book, for reference.) So adding additional copies of $A_k(V)$ for $k >n$ yields nothing new.

Alex Provost
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I think you are confusing covectors (elements of the dual space, or some tensor product of dual spaces) and elements of the exterior algebra on $V$ which I will denote $\Lambda^{k}V$.

You should try to prove that if the dimension of $V$ is $n$ then any element of $\Lambda^k V$ is zero for $k>n$ and then this means that the tail of the direct sum is a sum of trivial vector spaces, so is isomorphic to $\bigoplus_{k=0}^n{\Lambda^kV}$

Bernard W
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