This is a follow-up question for the answer: https://math.stackexchange.com/a/4214108/1140967.
In that answer, we see that there are three ways to associate a $k$-vector space $V$ with a $k$-functor:
- $X_V:=Spec(Sym(V^*))$;
- $Y_V:=Spec(Sym(V))$;
- $V_a$ the $k$-functor that sends a $k$-algebra $R$ to the $k$-vector space $V\otimes_k R$.
Question: If $V$ is infinite dimensional, is the functor $V_a$ representable by a $k$-scheme? If not, is $V_a$ an algebraic space?
What we know: We know that when $V$ is finite-dimensional, the functor $V_a$ is representable by $X_V$. We also know that when $V$ is infinite-dimensional, the functor $V_a$ is not representable by $X_V$.
Motivation: We can define the action of an algebraic group scheme $G$ on $V$ using either of the three $k$-functors above. When $V$ is finite-dimensional, no matter which of the three functions we use, an action of $G$ on $V$ is equivalent to a $G$-comodule. When $V$ is infinite-dimensional, the category of $G$-comodules is equivalent to the category of $G$-modules using $V_a$ (section 2.8 in Jantzen's Representations of Algebraic Groups; or Remark 4.1 of Milne's book), and is anti-equivalent to the category of $G$-modules using $Y_V$ (this is in the answer quoted above). Therefore, it seems that $V_a$ is really the natural "geometric" object associated with $V.$ I am thus interested to understand $V_a$ more geometrically.
