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I apologize in advance, but I need help with a rather simple question.

Given $L_k(V)$ is the set of all $k-$linear functions on the vector space $V$ on $\mathbb{R}$, what is $L_0(V)$? Does it make sense?

And why? I cannot simply get there from the definition. If $f\in L_k(V)$, $f$ is linear in all its arguments and is a map:

$f:V \times \dots \times V \rightarrow \mathbb{R}$

the cartesian product here has $k$ copies of $V$. How can I extrapolate to $k=0$? Thanks!

Giuseppe
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    So the Cartesian product has zero copies of $V$? For the empty Cartesian product see this post. – Dietrich Burde Nov 03 '22 at 14:02
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    I would guess the "product" in question should be the zero dimensional vector space. – kimchi lover Nov 03 '22 at 14:05
  • The question arose reading "An introduction to manifolds" by Tu. In the definition of the wedge product, the author mentions A0(V), the set of degree 0 alternating multilinear functions on V, and claims it is just real constants. – Giuseppe Nov 03 '22 at 14:06
  • @kimchilover This cannot be; the only linear map from a zero dimensional space is the trivial map, since $0$ has to map to $0$. What we want are linear maps $\R \to \R$. – Nicholas Todoroff Nov 03 '22 at 18:55

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$ \newcommand\R{\mathbb R} $

A multilinear function $V^{\times k} \to \R$ is characterized by the tensor power $V^{\otimes k} \to \R$, or put another way $(V^{\otimes k})^*$; the universal property of such a tensor power says that any multilinear function $f : V^{\times k} \to W$ for vector spaces $V, W$ extends uniquely to a linear map $f' : V^{\otimes k} \to W$ such that $$ f'(v_1\otimes v_2\otimes\cdots\otimes v_k) = f(v_1, v_2, \dotsc, v_k). $$ Note that this also means there is a one-to-one correspondence between multilinear functions and linear functions on $V^{\otimes k}$. So what we want to do is find $V^{\otimes 0}$, i.e. some vector space that satisfies an analogous universal property. The answer is $V^{\otimes 0} = \R$, for observe: every map (linear or not) $f : \{0\} \to W$ (which is just a choice of $f(0) \in W$) extends uniquely to a linear map $f' : \R \to W$ such that $f'(1) = f(0)$.

Thus the "multilinear maps of degree zero" are more properly elements of $(V^{\otimes0})^*$, which are linear maps $f : \R \to \R$, which amount to a choice of $x \in \R$ so that $f(y) = xy$.

Nicholas Todoroff
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