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Say I have an embedding: $f: X\to Y$ and a non-vanishing form $\omega$ on $Y$. Are there conditions on $f$ so that the pull-back $f^* \omega$ vanishes everywhere?

Or in a specific case, say $\omega$ is the non-trivial curvature form of some line bundle on $Y$, when does $f^* \omega$ vanish?

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1 Answer 1

I'm not sure if there is a general answer. I'll give two examples from symplectic geometry below. Your specific example on curvature forms seems like it should be a standard result but I don't know it off the top of my head; perhaps someone else can come by and shed some light on it.


Background: A symplectic manifold is a pair $(M, \omega)$ consisting of a smooth manifold $M$ and a closed, nondegenerate $2$-form $\omega \in \Omega^2(M)$; such forms care called symplectic forms. Symplectic manifolds are necessarily even-dimensional (in order for the nondegeneracy condition to possibly be satisfied). If $X$ is a smooth manifold and $$f: X \longrightarrow M$$ is an embedding, then $X$ is called a Lagrangian submanifold of $M$ if $f^\ast \omega = 0$ and $\dim(X) = \frac{1}{2} \dim(M)$. If $(M_1, \omega_1)$ and $(M_2, \omega_2)$ are two symplectic manifolds, a diffeomorphism $$\varphi: M_1 \longrightarrow M_2$$ is called a symplectomorphism if $\varphi^\ast \omega_2 = \omega_1$.


Example 1: If $X$ is any smooth manifold, the cotangent bundle $T^\ast X$ of $X$ has a natural symplectic form $\omega_{\text{can}} = - d\lambda$ where $\lambda$ is the $1$-form on $T^\ast X$ defined pointwise by $$\lambda_{(x,\xi)} = d\pi_{(x,\xi)}^\ast \xi$$ for any $(x, \xi) \in T^\ast X$, where $$\pi: T^\ast X \longrightarrow X$$ is the bundle projection. If $\alpha \in \Omega^1(X)$ is any $1$-form on $X$, then we can define an embedding of $X$ into $T^\ast X$ by $$f_\alpha: X \longrightarrow T^\ast X,$$ $$x \mapsto (x, \alpha_x).$$ Write $X_\alpha$ for the image of $X$ under $f_\alpha$.

Theorem. Let $\alpha \in \omega^1(X)$. Then $X_\alpha$ is a Lagrangian submanifold of $(T^\ast X,\omega_\text{can})$ if and only if $\alpha$ is closed.

See my answer here for the proof and more details. So here we see that $f^\ast_\alpha \omega_{\text{can}} = 0$ if and only if $d\alpha = 0$.


Example 2: Let $(M,\omega)$ be any symplectic manifold. Then $(M \times M,\omega \oplus -\omega)$ is a symplectic manifold, where $$\omega \oplus -\omega = \mathrm{pr}_1^\ast \omega - \mathrm{pr}_2^\ast \omega.$$ Let $$\varphi: M \longrightarrow M$$ be a diffeomorphism. Then we can define an embedding of $M$ into $M \times M$ by $$f_\varphi: M \longrightarrow M \times M,$$ $$x \mapsto (x, \varphi(x)).$$ Write $M_\varphi$ for the image of $M$ under $f_\varphi$.

Theorem. $\varphi$ is a symplectomorphism if and only if $M_\varphi$ is a Lagrangian submanifold of $(M \times M, \omega \oplus -\omega)$.

Proof. We have that \begin{align*} f_\varphi^\ast (\omega \oplus -\omega) & = f_\varphi^\ast \mathrm{pr}_1^\ast \omega - f_\varphi^\ast \mathrm{pr}_2^\ast \omega \\ & = (\mathrm{pr}_1 \circ f_\varphi)^\ast \omega - (\mathrm{pr}_2 \circ f_\varphi)^\ast \omega \\ & = \mathrm{Id}^\ast_M \omega - \varphi^\ast \omega \\ & = \omega - \varphi^\ast \omega. \end{align*} So $M_\varphi$ is Lagrangian if and only if $f_\varphi^\ast (\omega \oplus -\omega) = 0$, which is true if and only if $\omega - \varphi^\ast \omega = 0$, or in other words $$\omega = \varphi^\ast \omega,$$ so that $\varphi$ is a symplectomorphism.

So here we see that the pullback being identically zero depends on $\varphi$ preserving the symplectic form, which seems to be a fundamentally different reason than the previous example. This makes me believe that a general answer might not be possible.

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