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Suppose we have the following map:

$$(\Omega^1(\mathbb{R}^n))^3\longrightarrow(\Omega^2(\mathbb{R}^n))^3$$

$$(\alpha,\beta,\gamma)\longmapsto(\mathrm{d}\alpha+\beta\wedge\gamma,\mathrm{d}\beta+\gamma\wedge\alpha,\mathrm{d}\gamma+\alpha\wedge\beta)$$

Is it injective / surjective? Which is its kernel / cokernel? Does it depend on $n$?

Any suggestion is welcome.

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    $\begingroup$ The map vanishes just on the 1-forms that satisfy the structure equations of $SO(3)$ or are all zero. Therefore the manifold is locally a product of homogeneous spaces of $SO(3)$. $\endgroup$ – Ben McKay Mar 24 '15 at 7:17
  • $\begingroup$ @BenMcKay Yes, indeed this problem arises when working with $SO(3)$ (or $SU(2)$). The really critical point for me to prove is the suprajectivity... I'm convinced this map is onto, but it is quite involved... $\endgroup$ – Jjm Mar 24 '15 at 7:20
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Your map is not onto for $n>2$, even locally. If $(A,B,C)$ is a triple of $2$-forms on $\mathbb{R}^n$ that can be written in the form $$ (A,B,C) = \bigl(\mathrm{d}\alpha + \beta\wedge\gamma, \mathrm{d}\beta + \gamma\wedge\alpha, \mathrm{d}\gamma + \alpha\wedge\beta\bigr), $$ then, taking the exterior derivative of these equations, we find $$ (\mathrm{d}A,\mathrm{d}B,\mathrm{d}C) = \bigl(B\wedge\gamma-\beta\wedge C,\ C\wedge\alpha-\gamma\wedge A,\ A\wedge\beta-\alpha\wedge B\bigr). $$ In particular, if $A$, $B$, and $C$ vanish at a point $x\in\mathbb{R}^n$ at which $(\mathrm{d}A,\mathrm{d}B,\mathrm{d}C)$ does not vanish, then the $1$-forms $\alpha$, $\beta$, and $\gamma$ cannot exist on a neighborhood of $x$. (Such examples are trivial to construct.)

Moreover, when $n>4$, this map does not contain the generic triple $(A,B,C)$ in its image.

The cases $n=3$ and $n=4$ are special, and, for suitable genericity hypotheses, one can prove surjectivity in special cases and under the right conditions, but it is, indeed, somewhat delicate.

By the way, this is not a 'strange problem'. It is known as the problem of prescribed curvature for $\mathrm{SU}(2)$ connections. For more information, you might look at my answer to this question.

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  • $\begingroup$ Thank you very much. Yes, as you have guessed, this problem has arosen when dealing with curvatures! $\endgroup$ – Jjm Mar 24 '15 at 11:07

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