Justification for $\operatorname{dim} \wedge^k(V)$ is $n\choose k$ the alternating $k$-linear form.

abstract-algebraexterior-algebralinear algebraskew-symmetric matricestensor-products

I was told that: $\operatorname{dim} \wedge^k(V)$ is $n\choose k.$ and I am trying to justify this by computing it for different dimensions of $V.$

If $\operatorname{dim}(V) = 1,$ then $v_1 \wedge v_1 = 0$. Therefore, $\operatorname{dim} \wedge^1(V)$ is $1\choose 1$

If $\operatorname{dim}(V) = 2,$ then we will have the following cases:

\begin{align*}
v_1 \wedge v_1 &= 0\\
v_1 \wedge v_2 &= – v_2 \wedge v_1\\
v_2 \wedge v_1 &= – v_1 \wedge v_2\\
v_2 \wedge v_2 &= 0
\end{align*}

But then I can not conclude what is my $k$ in this case and how many variables actually I am choosing as I have the second case is just the same as the third case. Could anyone help me in this please?

Also, I know that

\begin{align*}
v_1 \wedge v_1 \wedge v_1 &= 0\\
v_1 \wedge v_2 \wedge v_3 &= – v_2 \wedge v_1 \wedge v_3\\
&= – v_3 \wedge v_2 \wedge v_1\\
&= – v_1 \wedge v_3 \wedge v_2\\
&= v_2 \wedge v_3 \wedge v_1\\
&= v_3 \wedge v_1 \wedge v_2\\
v_2 \wedge v_2 \wedge v_2 &= 0\\
v_3 \wedge v_3 \wedge v_3 &= 0
\end{align*}

But I do not know how to translate this into $n \choose k.$ Could someone help me in this also?

Thanks in advance!

EDIT:

Note that I was told that my question maybe a duplicate of this If $V$ is $k$-dimensional then show that $\dim \wedge^n V = \binom {k} {n}.$ but I do not agree with this as I need a small and concrete example to be calculated not an abstract procedure as in the previous link and the other one is asking about a specific case when $k<n$ in $k \choose n$ which is not my case.

Best Answer

If $v_1, \dots, v_n$ is a basis for $V$ then for

  • $k = 1$, we have $v_1, \dots, v_n$ as a basis for $\bigwedge^1V = V$ (definition)
  • $k = 2$, we have $\{v_i \wedge v_j : i < j\}$ as a basis for $\bigwedge^2 V$
  • $k = 3$, we have $\{v_i \wedge v_j \wedge v_l : i < j < l\}$ as a basis for $\bigwedge^3 V$.

and so forth. By definition, $k$ is the number of vectors you see in each expression. So $v_1$ is $k = 1$ since there is one vector in the expression; $v_1 \wedge v_2$ is $k = 2$ since there are two vectors, and so forth.

You said you understand that $v_i \wedge v_j = - v_j \wedge v_i$ and $v_i \wedge v_j \wedge v_l = v_j \wedge v_l \wedge v_i =$ etc. so I won't belabour that point.

Imagine $\dim V = 3$ and $k = 2$. Then a general vector in $V$ looks like $av_1 + bv_2 + cv_3$ and you can expand, for example $$(av_1 + bv_2 + cv_3) \wedge v_2 = a (v_1 \wedge v_2) + b (v_2 \wedge v_2) + c(v_3 \wedge v_2) = a (v_1 \wedge v_2) - c (v_2 \wedge v_3)$$

Using the identities $v_i \wedge v_i = 0$ and $v_j \wedge v_i = - v_i \wedge v_j$.

If you expand $(av_1 + bv_2 + cv_3) \wedge (a'v_1 + b'v_2 + c'v_3)$ and collect terms, you will get some linear combination of $v_1 \wedge v_2, v_1 \wedge v_3, v_2 \wedge v_3$. (First expand one side, then the other.)

Next there is the combinatorial identity: $\#\{(i, j) : i < j\} = \#\{\{i, j\} : i \neq j \} = \binom{n}{2}$ because given a set of numbers like $\{5, 3\}$ there is exactly one way to create an increasing sequence $3 < 5$ out of it. Therefore increasing sequences of length $k$ and subsets of size $k$ are equinumerous.

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