Let $$ V := \frac{x_4+i\vec{x}\cdot{\vec{\sigma}}}{\left|x\right|}$$ where $\left(x_1,x_2,x_3,x_4\right)\in\mathbb{R}^4$, $|x|$ is the Euclidean norm, and $\sigma^j$ are the Pauli matrices.

Let $\theta_1,\,\theta_2,\,\theta_3$ be some parametrization of $S^3$.

How do you "see" that \begin{align} \sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\varepsilon^{i j k} tr\left[V\left(\frac{\partial}{\partial \theta_i}V^{-1}\right)V\left(\frac{\partial}{\partial \theta_j}V^{-1}\right)V\left(\frac{\partial}{\partial \theta_k}V^{-1}\right)\right]\end{align} is proportional to $\left[\sin\left(\theta_1\right)\right]^2\sin\left(\theta_2\right)$, where $\varepsilon^{i j k}$ is the totally anti-symmetric tensor?

I have tried explicitly calculating this expression but after half a page of tediousness it seems like the wrong way to go. Using Mathematica I get the right result ($-12\times \left[\sin\left(\theta_1\right)\right]^2\sin\left(\theta_2\right)$), but I would like to know how to just "see" this result and compute the proportionality constant conveniently.

  • $\begingroup$ $V$ is an arbitrary $2 \times 2$ matrix spanned in the basis of Pauli matrices? Where did you see this formula? $\endgroup$ – cactus314 May 9 '14 at 20:42
  • $\begingroup$ @johnmangual: I guess, I saw it in a book called "Aspects of Symmetry" by Sidney Coleman on page 288 where he introduces the BPST instanton. $\endgroup$ – PPR May 9 '14 at 22:04
  • $\begingroup$ @Psycho_pr, it would be hepful if you could provide a clearer formulation of this in mathematical terms, e.g. in terms of unit quaternions. I take it your V is just a unit quaternion? I can't really follow the notation with partial derivatives. $\endgroup$ – Mikhail Katz May 11 '14 at 14:11
  • $\begingroup$ Hi, I don't really know how to notate it with quaternions (though I think it is done here hep1.c.u-tokyo.ac.jp/~kazama/QFT/instanton3-1.pdf ) but as for the partial derivatives, if I understand correctly (which may be false), you parametrize $S^3$ with three angles, $\theta_j$, and then clearly $x_j$ depend on those angles (generalized spherical coordinates), which is how you carry out the derivatives. $\endgroup$ – PPR May 11 '14 at 14:36
  • $\begingroup$ Could you replace the cryptic sum by its full expansion? I do not understand what you're trying to express. $\endgroup$ – WimC May 13 '14 at 14:14

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