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A multilinear form is a mapping

\begin{align} \Delta: V^n \rightarrow K \end{align}

where $V$ is a finite-dimensional vector space over field $K$.

It must meet the following requirements:

  • First:

\begin{align} &\Delta\left(a_1, \dots, a_{i-1}, a_i+a_i', a_{i+1}, \dots, a_n\right) \\ =\; &\Delta\left(a_1, \dots, a_{i-1}, a_i, a_{i+1}, \dots, a_n\right) \\ +\; &\Delta\left(a_1, \dots, a_{i-1}, a_i', a_{i+1}, \dots, a_n\right) \end{align}

  • Second:

\begin{align} &\Delta\left(a_1, \dots, a_{i-1}, \lambda a_i, a_{i+1}, \dots, a_n\right) \\ =\; \lambda&\Delta\left(a_1, \dots, a_{i-1}, a_i, a_{i+1}, \dots, a_n\right) \end{align}

I know that the multilinear forms form a vector space over $K$. Let $W$ be that vector space.

Now I want to figure out what $\dim_K W$ is but I don't know where to start. Can you help me?

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    Do you know about tensor products? – 2017-01-28
  • 0
    I don't know about tensor products. – 2017-01-28

2 Answers 2

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Hint: Define maps $\varphi_{i_1,\ldots,i_n}(v_1,\ldots,v_n) = (e_{i_1}^*v_1)\cdot (e_{i_2}^*v_2)\cdots (e_{i_n}^*v_n)$ where $\{e_i\}$ is base for $V$ and $\{e_i^*\}$ its dual base. Show that these are multilinear, linearly independent and generate your space.

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    @Marcel, let me know if you need more details. – 2017-01-28
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    I have now shown that the maps you've specified generate my vector space. Could you give me a little indication of how to show linear independence? – 2017-01-29
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    @Marcel, assume $\sum\alpha_I\varphi_I = 0$ (where $I = (i_1,\ldots,i_n)$) and evaluate it at $(e_{j_1},\ldots,e_{j_n})$. – 2017-01-29
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If you have a bilinear form $\varphi\colon U\times V\to K$, you can define a map $\varphi_v\colon U\to V^*$, for each $u\in U$, by $$ \varphi_u(v)=\varphi(u,v) $$ and this is a linear map.

Conversely, if you have a linear map $f\colon U\to V^*$, you can define a bilinear map $\hat{f}\colon U\times V\to K$ by $$ \varphi(u,v)=f(u)(v) $$

This implies that the space $\operatorname{Bilin}(U\times V,K)$ of bilinear maps $U\times V\to K$ has dimension $(\dim U)(\dim V^*)=(\dim U)(\dim V)$, since it is isomorphic to $\operatorname{Lin}(U,V^*)$.

More generally, $\operatorname{Bilin}(U\times V,W)$ is isomorphic to $$ \operatorname{Lin}(U,\operatorname{Lin}(V,W)) $$ where $W$ is any vector space. You can prove this in the same fashion.

We can generalize this to get an isomorphism $$\DeclareMathOperator{\ML}{Multilin} \ML(V_1\times V_2\times\dots\times V_n,W)\to \ML(V_1\times V_2\times\dots\times V_{n-1}, \operatorname{Lin}(V_n,W))\tag{*} $$ For each $x\in V_n$ and each $\varphi\in\ML(V_1\times V_2\times\dots\times V_n,W)$, define $$ \varphi_x\colon V_1\times V_2\times V_{n-1}\to \operatorname{Lin}(V_n,W)),\qquad \varphi_x(v_1,\dots,v_{n-1})=\varphi(v_1,\dots,v_{n-1},x) $$

For $n=1$, the dimension of $\ML(V_1,W)=\operatorname{Lin}(V_1,W))= (\dim V_1)(\dim W)$, so we can conjecture that $$ \dim\ML(V_1\times V_2\times\dots\times V_n,W)= (\dim V_1)(\dim V_2)\dotsm(\dim V_n)(\dim W) $$ and prove it by induction using the isomorphism in (*).