Suppose we have the following map:



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

Any suggestion is welcome.

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ Ben's comment is not quite right. What he probaby meant is that, if $(\alpha,\beta,\gamma)$ goes to zero under your map (which, since it is nonlinear, I wouldn't describe as 'being in the kernel'), then there is a smooth map $g:\mathbb{R}^n\to\mathrm{SO}(3)$ that satisfies $$\begin{pmatrix} 0 &-\gamma&\beta\\\gamma&0&-\alpha\\-\beta&\alpha&0\end{pmatrix} = g^{-1}\mathrm{d}g.$$ Such a $g$ is unique up to left multiplication by a constant element of $\mathrm{SO}(3)$. (This is sometimes known as Cartan's Fundamental Lemma for $\mathrm{SO}(3)$.) $\endgroup$ Jun 14 '20 at 9:08

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.