# How to show a vector space is a real form of another?

I only know from the definition that, say $$V_1$$ is a real form of $$V$$, if $$V=\mathbb{C}\otimes_\mathbb{R} V_1$$, but what does this really mean? Is it true it is just saying that $$V$$ is spanned by vectors of $$V_1$$ over $$\mathbb{C}$$?

For example, if I want to show that $$\mathfrak{su}(n)$$ is a real form of $$\mathfrak{sl}_n(\mathbb{C})$$, what exactly am I suppose to check?

Thanks for any help!

• There are many confusing, or potentially confusing items in what you write. First, $A$ is a real form of $B$ can mean "$B$ is isomorphic to the complexification of $A$". But furthermore if $A\subset B$, it can also mean the stronger: the inclusion $A\subset B$ induces an isomorphism from the complexification of $A$ to $B$. Second, you're talking about being real form as vector space, but most likely you want to have a real form as algebra, which is a stronger fact. – YCor Mar 7 at 13:42
• In any case, if you have a complex algebra $B$, a real subalgebra $A\subset B$ such that $B=A\oplus iA$, then you have the strongest version: the inclusion induces an isomorphism from the complexification to $B$. So all you need here is to show that $\mathfrak{sl}_n(\mathbf{C})=\mathfrak{su}(n)\oplus i\mathfrak{su}(n)$. Since the real dimension of $\mathfrak{su}(n)$ is half the real dimension of $\mathfrak{sl}_n(\mathbf{C})$, it's enough to check that $\mathfrak{su}(n)\cap i\mathfrak{su}(n)=\{0\}$, and this is very easy. – YCor Mar 7 at 13:45

By definition $$V = V_1 \otimes_{\Bbb R} \Bbb C$$ is a complex vector space whose elements are formal linear combinations of pairs $$(v, z) \in V \times \Bbb C$$ modulo the usual identifications of tensor products, including in particular the the equivalence relation $$(a v, z) \sim (v, a z)$$, $$a \in \Bbb R$$. We denote the image of a single pair $$(v, z)$$ under this related by $$v \otimes z$$. Then, if $$(E_1, \ldots, E_n)$$ is a basis of $$V_1$$, then $$(E_1 \otimes 1, \ldots, E_n \otimes 1)$$ is a basis of $$V_1 \otimes_{\Bbb R} \Bbb C$$, but often by abuse of notation we write this latter basis again as $$(E_1, \ldots, E_n)$$.
Given a complex vector space $$\Bbb V$$, one way to construct a real form thereof is to pick a complex-antilinear involution $$\phi : \Bbb V \to \Bbb V$$---here involution just means that $$\phi^2 = 1_{\Bbb V}$$. Then, the fixed point set $$\Bbb V^{\phi} = \{{\bf v} \in \Bbb V : \phi({\bf v}) = {\bf v}\}$$---which is a real vector space---is a real form of $$\Bbb V$$.
Conversely, all real forms arise this way: Given a real vector space $$\Bbb W$$, the conjugation map characterized by $$\phi({\bf w} \otimes z) := {\bf w} \otimes \bar z$$ is an antilinear involution of the complexification $$\Bbb W \otimes_{\Bbb R} \Bbb C$$, and its fixed point set $$(\Bbb W \otimes_{\Bbb R} \Bbb C)^{\phi}$$ is the (real) span of the vectors $${\bf w} \otimes 1$$. Respectively identifying these vectors with $${\bf w} \in \Bbb W$$ defines a natural isomorphism $$(\Bbb W \otimes_{\Bbb R} \Bbb C)^{\phi} \cong \Bbb W$$. Similarly, if the real vector space $$\Bbb W$$ is a real form of the complex vector space $$\Bbb V$$, then there is a (complex) vector space isomorphism $$\Bbb V \cong \Bbb W \otimes_{\Bbb R} \Bbb C$$.
For example, we usually identify $$\mathfrak{su}(n)$$ as the space of $$n \times n$$ matrices that (1) are skew-Hermitian, that is, satisfy $$X^\dagger = -X$$, and (2) have zero trace (over $$\Bbb C$$), that is, are in $$\mathfrak{sl}(n, \Bbb C)$$. Rearranging the condition in (1) gives that these matrices are precisely the matrices in $$\mathfrak{sl}(n, \Bbb C)$$ fixed by the map $$\phi : X \mapsto -X^\dagger$$. This is an antilinear involution, so $$\mathfrak{sl}(n, \Bbb C)^{\phi} = \mathfrak{su}(n)$$, that is, (the underlying vector space of) $$\mathfrak{su}(n)$$ is a real form of (the underlying vector space of) $$\mathfrak{sl}(n, \Bbb C)$$.
Remark In the previous paragraph we said that the vector space underlying one real Lie algebra was a real form of the vector space underlying a certain complex Lie algebra. This extra verbiage is a priori necessary, because in the context of Lie algebras we usually reserve the terminology real form for when the conjugation operation is also compatible with the Lie algebra structure. In terms of the antilinear map $$\phi$$, this additional requirement is that $$\phi$$ is a (real) Lie algebra homomorphism, that is, that $$\phi([X, Y]) = [\phi(X), \phi(Y)]$$. This condition holds for the map $$\phi$$ in the example in the previous paragraph, so we conclude that $$\mathfrak{su}(n)$$ is a real form of the $$\mathfrak{sl}(n, \Bbb C)$$ in this stronger sense, too.