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Let $f(x)=\left(I-2\frac{Wxx^TW^T}{\|Wx\|^2}\right)x$ for $x\in \mathbb{R}^n$ and $W\in\mathbb{R}^{n\times n}$.

Suppose I could constraint $W$ so the Jacobian determinant $\det \left(\frac{\partial f(x)}{\partial x}\right)\neq 0$ for all $x$ in the n-sphere $S^{n-1}=\{x\in\mathbb{R}^{n} \mid \|x\|=1\}$.

Is $f$ then invertible on $S^d$? If, can we compute or approximate the inverse somehow?

Alexander Mathiasen
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    Using the result described here, we can see that having the Jacobian determinant be non-zero for all $x$ is indeed a sufficient condition to ensure a global inverse. However, this result does not give us a formula for this inverse – Ben Grossmann Sep 07 '20 at 11:12
  • So this would require me to check that $S^d$ is smooth, connected and simply connected. Furthermore, $f$ must be $C^1$ (apparently $f$ is proper if $f$ is $C^1$ https://math.stackexchange.com/questions/3817213/proper-map-on-n-sphere). – Alexander Mathiasen Sep 07 '20 at 12:43
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    Yes, but this is easy. $S^d$ is always smooth and connected and is simply connected whenever $d \geq 2$. For all $x$ at which there is no division by zero, $f$ is $C^1$ (in fact, completely smooth i.e. $C^\infty$) because its components can be written completely in terms of addition, mutliplication, and division of variables. – Ben Grossmann Sep 07 '20 at 13:07
  • I see. I then conclude $f$ is invertible if non-zero Jacobian determinant. Let $y=f(x)$ and note we can write $y=H(Wx)x$ where $H(v)=I-2vv^T/||v||^2$ is a Householder matrix. Then $x=H(Wx)y=g(x)$. Since $g$ is continuous on $S^d$ the fixed point iteration $g(x), g(g(x)), ...$ should converge to a fixed point. But $f$ is invertible so this fixed point must be exactly $x$. Do you concur? – Alexander Mathiasen Sep 08 '20 at 08:59
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    If you want $y = H(Wx)x$, then you need to add a factor of $2$ to $f$. And no, the fixed point iteration should not converge except in the case that $x = H(Wx)x$. In particular, note that $g^{k}(x)$ (exponent here means $k$ iterations) is equal to $x$ whenever $k$ is even and $H(Wx)x$ whenever $k$ is odd. – Ben Grossmann Sep 08 '20 at 09:10
  • I forgot the factor $2$, very sorry. So $g^1(x)=g(x)=H(Wx)y$ and $g^2(x)=H(Wg^1(x))y=H(W H(Wx)y)y$. Why is $g^2(x)=x$? If so, two iterations should give us the inverse and we are done, no? – Alexander Mathiasen Sep 08 '20 at 09:18
  • Oh sorry, I misunderstood what you meant by $g(x)$. That said, there is no guarantee that the fixed point iteration will converge – Ben Grossmann Sep 08 '20 at 09:22

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