Let $M$ be a complex projective manifold with an antiholomorphic involution. Can $M$ be defined by equations with real coefficients then?



Complex projective manifold with an antiholomorphic involution
**Corrected.** As Robert Bryant points out, it is not enough to show that the manifold can be realised as a submanifold of $\mathbb CP^n$ invariant under some anti-holomorphic involution of $\mathbb CP^n$. For this reason the answer is extended.




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Let $\sigma: M\to M$ be the anti-holomorphic involution. Take a very ample divisor $D$ on $M$, and consider the linear system $|D+\sigma(D)|=\mathbb P\_{\mathbb C}^n$. It is invariant under $\sigma$ and $\sigma$ induces on it an anti-holomorphic involution. Now, we have the embedding $M\to \mathbb P\_{\mathbb C}^{n \*}$ that is invariant under the induced action of $\sigma$.




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However, being invariant under an anti-holomorphic involution of $\mathbb CP^n$ doesn't mean being defined by equations with real coefficients (as Robert points out).


So to answer the question, we need to solve additionally the following exercise:


**Exercise.** Suppose we have $\mathbb CP^n$ with an anti-holomorphic involution $\sigma$. Find an embedding $\mathbb CP^n\to \mathbb CP^N$, such that the image is invariant under the real involution of $\mathbb CP^N$ and the real involution induces $\sigma$ on $\mathbb CP^n$.


**Solution.** Let us show first how to do this for $\mathbb CP^1$ with the antipodal involution (the example proposed by Robert). Here we can embed it in $\mathbb CP^2$ with equation $x^2+y^2+z^2=0$.


As for general case we note the following. On an even-dimensional space each anti-holomoprhic involution is conjugate to the usual complex conjugation $(z\_1:\ldots :z\_{2n+1})\to (\bar z\_1:\ldots :\bar z\_{2n+1})$. So we don't need to do anything if $n=2m$. On the other hand, on odd dimensional projective spaces there is one more type of anti-holomorphic involutions, namely $$(z\_1:\ldots: z\_m:z\_{m+1}:\ldots: z\_{2m}\to (-\bar z\_{m+1}:\ldots: -\bar z\_{2m}: -\bar z\_1:\ldots: -\bar z\_m ).$$


So we just need to take some Segre embedding of $\mathbb CP^{2m-1}$ into an even dimensional projective space. For this we need to find $k$ such that ${2m+k-1}\choose{k}$ $-1$ is even. This can always be done.


