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Let $X$ be a 2-dimensional complex manifold and let $Y$ be a smooth submanifold of $X$ of two real dimension. Let $J$ be the complex structure on the tangent space $T_xX$. The submanifold $Y$ is called totally real if $T_xY\cap J(T_xX)=\{0\}$ for all $x\in Y$.

Among the following submanifolds $Y$: $S^2$, $S^1\times S^1$ and $\mathbb RP^2$, which one can be embedded in $X$ as a totally real submanifold. Why $\mathbb RP^2$ can't be one of them?

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    $\Bbb{RP}^2 \subset \Bbb{CP}^2$ is totally real. Something is missing in your assumptions. –  Nov 18 '18 at 20:53
  • Thanks, I will edit it now –  Nov 18 '18 at 20:57
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    The affine quadric is symplectomorphic to the cotangent bundle of $S^{2}$ (see section 2.1 here https://arxiv.org/pdf/1012.4146.pdf), so there is a lagrangian sphere by the Weinstein N-hood theorem. – Nick L Nov 18 '18 at 21:40

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It depends on $X$, but all $3$ of those surfaces can be embedded as totally real submanifolds in some complex manifold of dimension $2$.

In all cases we can use the fact that for a lagrangian submanifold in a Kähler manifold is totally real (which is proven in the answer to this question What is the relation between totally real submanifold and Lagrangian submanifold?).

  1. $S^{1} \times S^{1}$ is Lagrangian in $\mathbb{CP}^{2}$ since it is a toric manifold, and the pre-image of a regular value of the toric moment map is a lagrangian torus.

  2. $\mathbb{RP}^{2}$ is a lagrangian in $\mathbb{CP}^{2}$, embedded by just taking points whose homogeneous co-ords are real numbers.

  3. To see $S^{2}$ as a lagrangain, just take a smooth rational curve with self-intersection $-2$ in a del Pezzo surface, which can be achieved by blowing up a point and then a point on its strict transform.

Nick L
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  • Thanks. I forgot to mention that $X$ is either $SL_2(\mathbb C)/\mathbb C^*$ the affine quadric or $\mathbb CP^1\times\mathbb C P^1$. Can the real projective space be embedded in $X$ in those cases? –  Nov 18 '18 at 21:05
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    @AmratA Note that the affine quadric is an open subset of $\Bbb{CP}^1 \times \Bbb{CP}^1$, not the whole thing. Once you specify $X$ this becomes harder to answer. –  Nov 18 '18 at 21:16
  • @Nick L, I see now the relation between Lagrangian submanifolds and totally real submanifolds. Does 3. in your answer show that $S^2$ is a Lagrangian submanifold of $\mathbb CP^2$?! –  Nov 20 '18 at 23:20
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    @AmratA No, it shows $S^2$ as a Lagrangian submanifold of something much more complicated. I do not think it is true that $S^2$ may be embedded as a Lagrangian submanifold of $\Bbb{CP}^2$, but I am not well-versed in symplectic geometry, so I will let someone more expert confirm. –  Nov 21 '18 at 05:47
  • Thanks Mike. I have asked this in a separate question. –  Nov 21 '18 at 06:00