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In W. Tu's An Introduction to Manifolds, the following definition is given:

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

I am wondering why the second equality holds.

If I have $f(v_1,\dots,v_n)=0$, which is clearly alternating and $n$-linear, so $f\in A_n(V)$. Then $a=f\wedge f\in A_{2n}(V)$, which implies $\displaystyle a\in\bigoplus_{k=0}^\infty A_k(V)$; but $a$ cannot be represented uniquely by a finite sum $a_{i_1}+\cdots+a_{i_m}$ where $a_{i_j}\in A_{i_j}(V)$, $0\leq i_j\leq n$ ($a$ is $2n$-linear, but all the $a_{i_j}$ are at most $n$-linear), thus $a$ is not an element of $\displaystyle\bigoplus_{k=0}^n A_k(V)$.


In Tu's book, $A_k(V)$ represents the set of all alternating $k$-linear functions with domain $V^k$ and codomain $\mathbb{R}$; $\bigoplus$ is the symbol for direct sum of vector spaces.

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This is because if $n=\dim V$, then every alternating $k$-linear form is zero if $k>n$.

So there is some abuse of notation going on. $$A_*(V)=\bigoplus_{k=0}^\infty A_k(V)=\bigoplus_{k=0}^nA_k(V) \oplus \bigoplus_{k=n+1}^\infty A_k(V),$$ its just that for $k>n$ we have $A_k(V)=\{0_{A_k}\},$ so the second summand is omitted in the formula.

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@abdul Thanks for the answer; now I understand. Here are some of my supplementals regarding your first sentence.

Claim. For $f\in A_k(V)$, if for $i\neq j$ we have $v_i=v_j$, then $f(\dots,v_i,\dots,v_j,\dots)=0$.

Proof. Assume WLOG that $i<j$. Define a permutation such that $\sigma(i)=j$ and $\sigma(j)=i$; we claim that $\operatorname{sgn}\sigma=-1$. This can be shown by observing the fact that moving $v_i$ to $j$-th position requires $(j-i)$ transpositions and then $v_j$ (at position $(j-1)$ to $i$-th requires $(j-i-1)$. Clearly $(j-i)+(j-i-1)=2(j-i)-1$ is always an odd number, thus $\operatorname{sgn}\sigma=-1$ and $f=-f$; since the codomain of $f$ is $\mathbb{R}$, $f=0$.

Claim. For $f\in A_k(V)$, if $(v_1,\dots,v_k)$ is linearly dependent, then $f(v_1,\dots,v_k)=0$.

Proof. By the linear dependence of the group, some $v_i$ of the vectors can be represented by the linear combination of others, i.e. $$v_i=\alpha_1v_1+\dots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\dots+\alpha_kv_k.$$ Use this to expand $f$ by linearity $$f(\dots,v_i,\dots)=\alpha_1f(v_1,\dots,v_1,\dots)+\dots+\alpha_kf(\dots,v_k,\dots,v_k),$$ which, by the previous claim, is $0$.

For $f\in A_k(V)$ with $k>\dim V$, every possible group of parameters of $f$ is linearly independent, then by the previous claim $f(a)=0$ for all $a\in V^k$, thus $f=0$.

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