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The tangent bundle $TM$ of a Riemannian manifold M has a natural Kähler structure with the Kähler form agreeing with the canonical symplectic form of $TM$ coming from the cotangent bundle.

To see this, pick local coordinates $\mathbf{x}=(x_1,\ldots,x_n)$ on M and let the metric be given by a positive definite matrix A $$g = d\mathbf{x}^TAd\mathbf{x}$$ Introduce complex coordinates $\mathbf{z}=\mathbf{x}+i\ d{\mathbf{x}}$ and lift the the metric to a Hermitian metric $h$ on $TM$ $$h = d\mathbf{z}^*Ad\mathbf{z} = (d\mathbf{x}-i\ d^2\mathbf{x})^T\ A\ (d\mathbf{x}+i\ d^2\mathbf{x})$$ (Here $d^2\mathbf{x}=(d^2x_1,\ldots,d^2x_n)$ are coordinates on the second order tangent space.)

The Kähler form is $$\Omega = -\text{Im}\ h(d\mathbf{z}_1, d\mathbf{z}_2) = d^2\mathbf{x}_1^T\ A\ d\mathbf{x}_2 - d\mathbf{x}_2^Td\mathbf{x}_1^T\ A\ d^2\mathbf{x}_1 d^2\mathbf{x}_2$$

and since the momentums (cotangent coordinates) are $\mathbf{p}=d\mathbf{x}^TA$, the Kähler form becomes $$\Omega = d\mathbf{p}_1^Td\mathbf{p}_1\ d\mathbf{x}_2 - d\mathbf{p}_2^Td\mathbf{p}_2\ d\mathbf{x}_1$$ which is the canonical symplectic form of $T^*M$.

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The tangent bundle $TM$ of a Riemannian manifold M has a natural Kähler structure with the Kähler form agreeing with the canonical symplectic form of $TM$ coming from the cotangent bundle.

To see this, pick local coordinates $\mathbf{x}=(x_1,\ldots,x_n)$ on M and let the metric be given by a positive definite matrix A $$g = d\mathbf{x}^TAd\mathbf{x}$$ Introduce complex coordinates $\mathbf{z}=\mathbf{x}+i\ d{\mathbf{x}}$ and lift the the metric to a Hermitian metric $h$ on $TM$ $$h = d\mathbf{z}^*Ad\mathbf{z} = (d\mathbf{x}-i\ d^2\mathbf{x})^T\ A\ (d\mathbf{x}+i\ d^2\mathbf{x})$$ (Here $d^2\mathbf{x}=(d^2x_1,\ldots,d^2x_n)$ are coordinates on the second order tangent space.)

The Kähler form is $$\Omega = -\text{Im}\ h(d\mathbf{z}_1, d\mathbf{z}_2) = d^2\mathbf{x}_1^T\ A\ d\mathbf{x}_2 - d\mathbf{x}_2^T\ A\ d^2\mathbf{x}_1$$

and since the momentums (cotangent coordinates) are $\mathbf{p}=d\mathbf{x}^TM$, \mathbf{p}=d\mathbf{x}^TA$, the Kähler form becomes $$\Omega = d\mathbf{p}_1^T\ d\mathbf{x}_2 - d\mathbf{p}_2^T\ d\mathbf{x}_1$$ which is the canonical symplectic form of$T^*M$. 1 The tangent bundle$TM$of a Riemannian manifold M has a natural Kähler structure with the Kähler form agreeing with the canonical symplectic form of$TM$coming from the cotangent bundle. To see this, pick local coordinates$\mathbf{x}=(x_1,\ldots,x_n)$on M and let the metric be given by a positive definite matrix A $$g = d\mathbf{x}^TAd\mathbf{x}$$ Introduce complex coordinates$\mathbf{z}=\mathbf{x}+i\ d{\mathbf{x}}$and lift the the metric to a Hermitian metric$h$on$TM$$$h = d\mathbf{z}^*Ad\mathbf{z} = (d\mathbf{x}-i\ d^2\mathbf{x})^T\ A\ (d\mathbf{x}+i\ d^2\mathbf{x})$$ (Here$d^2\mathbf{x}=(d^2x_1,\ldots,d^2x_n)$are coordinates on the second order tangent space.) The Kähler form is $$\Omega = -\text{Im}\ h(d\mathbf{z}_1, d\mathbf{z}_2) = d^2\mathbf{x}_1^T\ A\ d\mathbf{x}_2 - d\mathbf{x}_2^T\ A\ d^2\mathbf{x}_1$$ and since the momentums (cotangent coordinates) are$\mathbf{p}=d\mathbf{x}^TM$, the Kähler form becomes $$\Omega = d\mathbf{p}_1^T\ d\mathbf{x}_2 - d\mathbf{p}_2^T\ d\mathbf{x}_1$$ which is the canonical symplectic form of$T^*M\$.