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Ali Taghavi
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A two$2$- form $\alpha$ on a n- manifold $M$ is called anti symplectic if for every $x\in M$, $\{ v\in T_{x} M \mid i_{v} \alpha=0 \}$ is a $n-2$ dimensional subspace of $T_{x}M$. So we obtain a $n-2$ dimensional subbundle of $T M$. We denote it by $\tilde{\ker} \;\alpha$

The first question:

Under what condition on $\alpha$, the distribution $\tilde{\ker} \;\alpha$ is integrable? (I am motivated by Frobenious condition $\alpha \wedge d\alpha=0$, so I search for an algebraic condition)

The second question:

Assume that $\omega$ is a symplectic 2-form on a 2n- manifold. Can we write $\omega$ in the global form $\omega=\sum_{i=1}^{n} \alpha_{i}$ where each $\alpha_{i}$ is anti symplectic form.(I think that the local argument and then using partition of unity, does not work)

A two form $\alpha$ on a n- manifold $M$ is called anti symplectic if for every $x\in M$, $\{ v\in T_{x} M \mid i_{v} \alpha=0 \}$ is a $n-2$ dimensional subspace of $T_{x}M$. So we obtain a $n-2$ dimensional subbundle of $T M$. We denote it by $\tilde{\ker} \;\alpha$

The first question:

Under what condition on $\alpha$, the distribution $\tilde{\ker} \;\alpha$ is integrable? (I am motivated by Frobenious condition $\alpha \wedge d\alpha=0$, so I search for an algebraic condition)

The second question:

Assume that $\omega$ is a symplectic 2-form on a 2n- manifold. Can we write $\omega$ in the global form $\omega=\sum_{i=1}^{n} \alpha_{i}$ where each $\alpha_{i}$ is anti symplectic form.(I think that the local argument and then using partition of unity, does not work)

A $2$- form $\alpha$ on a n- manifold $M$ is called anti symplectic if for every $x\in M$, $\{ v\in T_{x} M \mid i_{v} \alpha=0 \}$ is a $n-2$ dimensional subspace of $T_{x}M$. So we obtain a $n-2$ dimensional subbundle of $T M$. We denote it by $\tilde{\ker} \;\alpha$

The first question:

Under what condition on $\alpha$, the distribution $\tilde{\ker} \;\alpha$ is integrable? (I am motivated by Frobenious condition $\alpha \wedge d\alpha=0$, so I search for an algebraic condition)

The second question:

Assume that $\omega$ is a symplectic 2-form on a 2n- manifold. Can we write $\omega$ in the global form $\omega=\sum_{i=1}^{n} \alpha_{i}$ where each $\alpha_{i}$ is anti symplectic form.(I think that the local argument and then using partition of unity, does not work)

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Ali Taghavi
  • 356
  • 8
  • 31
  • 123
added 118 characters in body; edited tags
Source Link
Ali Taghavi
  • 356
  • 8
  • 31
  • 123

A two form $\alpha$ on a n- manifold $M$ is called anti symplectic if for every $x\in M$, $\{ v\in T_{x} M \mid i_{v} \alpha=0 \}$ is a $n-2$ dimensional subspace of $T_{x}M$. So we obtain a $n-2$ dimensional subbundle of $T M$. We denote it by $\tilde{\ker} \;\alpha$.

The first question:

Under what condition on $\alpha$, the distribution $\tilde{\ker} \;\alpha$ is integrable? (I am motivated by Frobenious condition $\alpha \wedge d\alpha=0$, so I search for an algebraic condition)

The second question:

Assume that $\omega$ is a symplectic 2-form on a 2n- manifold. Can we write $\omega$ in the global form $\omega=\sum_{i=1}^{n} \alpha_{i}$ where each $\alpha_{i}$ is anti symplectic form.(I think that the local argument and then using partition of unity, does not work)

A two form $\alpha$ on a n- manifold $M$ is called anti symplectic if for every $x\in M$, $\{ v\in T_{x} M \mid i_{v} \alpha=0 \}$ is a $n-2$ dimensional subspace of $T_{x}M$. So we obtain a $n-2$ dimensional subbundle of $T M$. We denote it by $\tilde{\ker} \;\alpha$.

The first question:

Under what condition on $\alpha$, the distribution $\tilde{\ker} \;\alpha$ is integrable? (I am motivated by Frobenious condition $\alpha \wedge d\alpha=0$, so I search for an algebraic condition)

The second question:

Assume that $\omega$ is a symplectic 2-form on a 2n- manifold. Can we write $\omega$ in the global form $\omega=\sum_{i=1}^{n} \alpha_{i}$ where each $\alpha_{i}$ is anti symplectic form.(I think that the local argument and then using partition of unity, does not work)

A two form $\alpha$ on a n- manifold $M$ is called anti symplectic if for every $x\in M$, $\{ v\in T_{x} M \mid i_{v} \alpha=0 \}$ is a $n-2$ dimensional subspace of $T_{x}M$. So we obtain a $n-2$ dimensional subbundle of $T M$. We denote it by $\tilde{\ker} \;\alpha$

The first question:

Under what condition on $\alpha$, the distribution $\tilde{\ker} \;\alpha$ is integrable? (I am motivated by Frobenious condition $\alpha \wedge d\alpha=0$, so I search for an algebraic condition)

The second question:

Assume that $\omega$ is a symplectic 2-form on a 2n- manifold. Can we write $\omega$ in the global form $\omega=\sum_{i=1}^{n} \alpha_{i}$ where each $\alpha_{i}$ is anti symplectic form.(I think that the local argument and then using partition of unity, does not work)

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Ali Taghavi
  • 356
  • 8
  • 31
  • 123
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