Real forms of complex vector spaces and $\mathbb{C}$-algebra

Question

A real form $W$ of a complex vector space $V$ is a real subspace s.t. $\mathbb{C}\otimes_{\mathbb{R}}W \cong V$ by $a\otimes x \longrightarrow ax$, or equivalently there is an $\mathbb{R}$-basis of $W$ that is also a $\mathbb{C}$-basis of $V$.

There is a fact that every complex vector space has a real form.

For example, $\mathbb{R}^n$, $\mathbb{R}[x]$ and $M_n(\mathbb{R})$ are real forms of $\mathbb{C}^n$, $\mathbb{C}[x]$ and $M_n(\mathbb{C})$ respectively. The last two examples are not only real vector spaces but real subalgebras.

So, I'm wondering whether any complex associative algebra $A$ has a real form, i.e. a real subalgebra $B$ s.t. $\mathbb{C}\otimes_{\mathbb{R}}B \cong A$ as $\mathbb{C}$-algbras? Is there any counterexample?

Answer 1

I am very sure that there are counterexamples, but haven't found one so far. In fact, I have found many examples which surprisingly have a real form. I share them here so that others interested in this problem won't have to fall for the same traps. I assume any algebra to be unital.

1. If $G$ is any monoid, then the monoid algebra $\mathbb{C}[G]$ has a real form. This applies in particular to free commutative algebras (i.e. polynomial rings in any set of variables) as well as to free algebras.
2. The class of $\mathbb{C}$-algebras with a real form is closed under arbitrary products and tensor products. Notice that although tensoring with a field extension doesn't have to preserve infinite products, this is the case for finite field extensions such as $\mathbb{C}/\mathbb{R}$.
3. A $\mathbb{C}$-algebra $A$ has a real form if and only if there is a homomorphism of $\mathbb{R}$-algebras ${}^* : A \to A$ which is an involution (i.e. $a=(a^*)^*$) and satisfies $(\lambda a)^* = \overline{\lambda} a^*$. Notice the similarity to the notion of a *-algebra, the difference is that we demand $(ab)^* = a^* b^*$. Proof: "$\Rightarrow$" On $\mathbb{C} \otimes_{\mathbb{R}} B$ define $(\lambda \otimes b)^* := \overline{\lambda} \otimes b$. "$\Leftarrow$" Consider the real subalgebra $B=\{a \in A : a^*=a\}$. Every $a \in A$ has a unique representation as $a=r + is$ with $r,s \in B$, namely $r = \frac{a+a^*}{2}$ and $s=\frac{a-a^*}{2i}$. Thus $\mathbb{C} \otimes_{\mathbb{R}} B \cong A$. $\square$
4. If $X$ is any topological space, then $C(X,\mathbb{C})$ has a real form, namely $C(X,\mathbb{R})$. In fact, we have the involution given by taking the complex conjugate pointwise.
5. If $D \subseteq \mathbb{C}^k$ is a (Edit: symmetric) domain, then the $\mathbb{C}$-algebra of holomorphic functions $D \to \mathbb{C}$ has an involution and therefore a real form: Apply the complex conjugation to the coefficients of a power series expansion.
6. If $B$ is any set of variables, then the function field $\mathbb{C}(B)$ has a real form. More generally, if $L/K$ is an algebraic field extension, then $L \otimes_K K(B) \cong L(B)$.
7. If $f \in \mathbb{C}[x]$, then $\mathbb{C}[x]/(f)$ has a real form. In fact, this algebra is isomorphic to $\prod_i \mathbb{C}[x]/(x^{k_i})$ if $k_i$ is the multiplicity of the $i$th root of $f$, and $\mathbb{C}[x]/(x^{k_i}) = \mathbb{C} \otimes_{\mathbb{R}} \mathbb{R}[x]/(x^{k_i})$.
8. I believe that there must be a polynomial $f \in \mathbb{C}[x,y]$ such that $\mathbb{C}[x,y]/(f)$ has no real form. In the language of algebraic geometry, this means that the complex curve $V(f)$ is not definable over $\mathbb{R}$. In the literature I could only find information about projective curves. So let's take a random polynomial (honestly, I just make it up while writing this) $f = x^3 + i x y + y^2 x + i$. Substituting $y$ with $iy$ gives $f' = x^3-xy - y^2 x + i$. Substituting $x$ with $ix$ gives $f''=-i x^3-ixy - iy^2 x + i$. Thus, $\mathbb{R}[x,y]/(-x^3-xy-y^2 x + 1)$ is a real form. This kind of substitution has worked in all random examples that I've tried.

Answer 2

Here is a sketch of a construction:

Let $k$ be a perfect field. The $j$-invariant provides a map $$j : \{\text{elliptic curves over } k\}/~{\cong} \to k$$ which is bijective when $k$ is algebraically closed. It commutes with base change (since the $j$-invariant can be computed from the Weierstrass equation). Now take an elliptic curve over $\mathbb{C}$ with $j$-invariant in $\mathbb{C} \setminus \mathbb{R}$. Its function field is a $\mathbb{C}$-algebra which is not induced by an $\mathbb{R}$-algebra, since this would have to be the function field of an elliptic curve over $\mathbb{R}$ which induces the given one over $\mathbb{C}$. But then it would have $j$-invariant in $\mathbb{R}$, a contradiction.