Timeline for On nowhere zero self-dual 2-forms
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| Jul 10, 2017 at 13:58 | comment | added | Robert Bryant | You are confusing various decompositions of the vector bundles involved. For example, $\mathrm{d}\omega$ is a real 3-form (since $\omega$ is a real $2$-form), and, as such, $\mathrm{d}\omega$ does not lie in either $\Lambda^{2,1}$ or $\Lambda^{1,2}$, but 'diagonally' in the sum of the two, as $\Lambda^3(T^*)\otimes\mathbb{C} = \Lambda^{2,1} \oplus \Lambda^{1,2}$. (In fact, each of the terms in your 'formula' for $\mathrm{d}\omega$ are sections of $\Lambda^3(T^*)$.) | |
| Jul 10, 2017 at 13:37 | comment | added | Varun | My apologies. Let me be more specific. The almost complex structure give the isomorphism $$ T^{\ast}X \otimes_{\mathbb{R}} \overline{K} \cong T^{\ast}X \otimes_{\mathbb{C}} K \oplus T^{\ast}X \otimes_{\mathbb{C}} \overline{K}. $$ where $K$ is a complex line bundle. Moreover, the wedge product gives a complex, bi-linear map $$ T^{\ast}X \times T^{\ast}X \longrightarrow \Lambda^2 T^{\ast}X = K. $$ using which, we can identify $TX \cong T^{\ast}X \otimes_{\mathbb{C}} \overline{K}$. These are the 2 components of $\nabla \omega$ which I mention in my question. | |
| Jul 10, 2017 at 12:16 | comment | added | Robert Bryant | Your assertion that $\nabla\omega$ can be interpreted as a section of $\Lambda^{1,2}\oplus \Lambda^{2,1}$ is in error. In fact, $\nabla\omega$ is a section of $\Lambda^2_+\otimes \Lambda^1$, a vector bundle of (real) rank $3\times 4 = 12$, whereas $\Lambda^{1,2}$ and $\Lambda^{2,1}$ are vector bundles of (real) rank $4$. You are missing another $4$-dimensional representation of $\mathrm{U}(2)$. By the way, as a representation of $\mathrm{SO}(4)$, $\Lambda^2_+\otimes \Lambda^1$ splits as the sum of two vector bundles, one of rank $4$ and the other of rank $8$. | |
| Jul 10, 2017 at 11:30 | history | asked | Varun | CC BY-SA 3.0 |