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I have never worked with differential geometry over $\mathbb{C}$, and I feel a little bit confused.

If $(M,J)$ is an almost complex manifold of dimension $2n$, we can extend $J_{p}$ to $T_{p}(M)\otimes_{\mathbb{R}}\mathbb{C}$ (which is a $2n$ dimensional $\mathbb{C}$-vector space), by $$J_{p}(v\otimes z)=J_{p}(v)\otimes z.$$ Then, $J_{p}$ is a $\mathbb{C}$-linear map and $J_{p}^{2}=-\text{Id}$, so $J_{p}$ has two eigenvalues: $i$ and $-i$

If $T_{1,0}$ is the $i$-eigenspace and $T_{0,1}$ is the $-i$-eigenspace (which are complex subspaces), it can be proved that $TM\otimes_{\mathbb R}\mathbb{C} \cong T_{1,0}\oplus T_{0,1}$.

Similarly, $T^{*}M\otimes_{\mathbb R} \mathbb{C} \cong T^{1,0}\oplus T^{0,1}$. Then, $\bigwedge^{k}(T^{*}M\otimes_{\mathbb{R}} \mathbb{C})$ equals $\oplus_{m+l=k}(\bigwedge^{m}(T^{1,0})\otimes \bigwedge^{l}(T^{0,1}))$.

My question is: I believe that the last tensor product should be taken over $\mathbb{C}$, since $T^{1,0}$ and $T^{0,1}$ are vector spaces over $\mathbb{C}$. Am I right?

Moreover, how do elements of $\bigwedge^{m}(T^{1,0})\otimes \bigwedge^{l}(T^{0,1})$ look like?

I am following these notes (page 78): https://people.math.ethz.ch/~acannas/Papers/lsg.pdf

Thanks in advance!

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    $\begingroup$ This question has been asked and answered many times on this site. Have you searched for your precise title? $\endgroup$ – Ted Shifrin Feb 16 '18 at 0:11
  • $\begingroup$ Yes, I have understood thanks to some answers the construction of $T_{p}M\otimes \mathbb{C}$ as a $\mathbb{C}$-vector space, but I still don't understand the exterior algebra, I have not found any question about that... $\endgroup$ – Laura Feb 16 '18 at 0:55

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