# Why is it that $\displaystyle\bigoplus_{k=0}^\infty A_k(V)=\bigoplus_{k=0}^{\dim V}A_k(V)$?

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.

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.

@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 $$i. Define a permutation such that $$\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 $$f=0$$.