What is the empty tensor product of vector spaces?

Question

The tensor product of a space with itself once is $V^{\otimes1}$, but what is $V^{\otimes0}$? Since it is an empty tensor product, it is - a fortiori - an empty product. So I'm looking for a "$1$" of some sort, just not sure what that would mean in this context.

"If I take the tensor product of a vector space with itself zero times, I would get ...",

and I am guessing here, but is it the underlying field, $\mathbb{F}$?

Answer 1

You are correct. In any case one has a notion of "product" and a "unit" for that product, the proper convention is that the "empty product" equals the "unit". If we are working with vector spaces over a field $F$ and the corresponding tensor product $\otimes_F$, the corresponding unit is the $F$-vector space $F$ itself. Indeed, there is a canonical isomorphism $V\otimes_FF\rightarrow V$ for any $F$-vector space $V$ given by $v\otimes\lambda\mapsto\lambda x$ (and similarly in the other order). Thus, the empty tensor product is $V^{\otimes0}=F$.

Answer 2

$ \newcommand\F{\mathbb F} \newcommand\Tensor{\mathop{\textstyle\bigotimes}} $Tensor powers are characterized as the "universal source" of multilinear maps. If $f : V^k \to W$ is multilinear then there is a unique linear extension $f'$ of $f$ to $V^{\otimes k} \to W$, with "extension" meaning $$f'(v_1\otimes\dotsb\otimes v_k) = f(v_1, \dotsc, v_k).$$ It will be important to be more precise. "The $k^{\text{th}}$-tensor power" is more properly a function $\iota$ with domain $V^k$, codomain a vector space $V^{\otimes k}$, and which satisfies the above universal property. In the notation above, $$ \iota(v_1,\dotsc,v_k) = v_1\otimes\dotsb\otimes v_k. $$

We need to give meaning to a "multilinear map $V^0 \to W$".

We could say $V^0 = \varnothing$; a multilinear map is a map that is linear in each of its arguments, so if the are no arguments then it is automatically multilinear. Then $f$ is the unique map $\varnothing \to W$ and also $\iota : \varnothing \to V^{\otimes0}$. This puts no constraints on $f'$. Because $f'$ has to be unique, our only choice is $V^{\otimes0} = \{0\}$.

If we (perhaps more reasonably) interpret $V^0$ as a Cartesian power then $V^0 = \{*\}$ is a one-element set and $f(*)$ is an arbitrary element of $W$. ($f$ doesn't have to be linear in any sense.) We also have $\iota(*)$ as an arbitrary element of $V^{\otimes0}$, and we have the constraint $f'(\iota(*)) = f(*)$. If $\iota(*) = 0$, then $f(*) = 0$; but $f(*)$ is arbitrary, so this can only happen when $W = \{0\}$ and so $V^{\otimes0}$ is unconstrained by this case. If $\iota(*) \ne 0$, then for $f'$ to be uniquely determined we need $V^{\otimes0} = \mathrm{span}\{\iota(*)\} \cong \F$.

---

The latter definition is favoured for a simple reason: $$ \F\otimes V \cong V\quad\text{but}\quad \{0\}\otimes V \cong \{0\}. $$

Answer 3

A tensor product of spaces corresponds to a Cartesian product of bases: if $\forall j: V_j = \operatorname{span} {\hat V}_j$, then $\bigotimes_j V_j = \operatorname {span} \prod_j {\hat V}_j$ (up to isomorphism). So you may equivalently look for the nullary Cartesian product: conventionally, this is a singleton of the 0-tuple. Therefore, the nullary tensor product is a one-dimensional space, isomorphic to the underlying field.

This answer is somewhat flawed in that it relies on assigning a basis to each vector space, which is an arbitrary choice (and one not always possible to make in the absence of a choice axiom). But it’s a useful intuition to have.