I am looking at the Hopf-fibration and I am looking at the preimage of the equator in $\mathbb{S}^2$. I think that I have proved that this is just the flat torus and now I want to calculate the principal curvatures of this torus. My general approach to the problem has been:

I consider $\mathbb{R}^4$ as $\mathbb{C}^2$ and $\mathbb{S}^3$ to be "the unit circle" in this "plane", every point on the circle determines a line through the origion. And the lines through the origion intersects the sphere in a unique circle. Mapping $\mathbb{S^3}$ to the space of lines through the origion gives me an onto map to $\mathbb{C}\mathbb{P}^1$ which I diffeomorphically identify with $\mathbb{S}^2$. The preimage of any point $(z_1:z_2)$ in $\mathbb{S}^3$ is just the circle $\frac{z_1}{z_2}=z$. Hence the preimage of the equator is all circles satisfying $\frac{|z_1|}{|z_2|}=1$, i.e $|z_1|=|z_2|$, hence the preimage is a product of two circles, i.e a torus.

Since my torus is just $S^1\times S^1$ in $\mathbb{C}\times \mathbb{C}$ it is equipped with the product metric, and hence in this case flat.

I am now stuck, here is two ways I would like to go forward:

Calculating the Weingarten map. Using the parametrization $(e^{i\theta },e^{i\psi })$, what is the normal direction inside of $\mathbb{S}^3$?

I have also tried to look at the normal curvature of curves in $S^1\times S^1$ are these just "circles wrapping around the torus" the smaller ones curving more then the bigger? This way of thinking would give me principal curvature $1$ and something depending on where on the torus I am being $-1$ in "the inner circle"?


One (of the possible two ones) unit normal vector of $\mathbb{S}^1 \times \mathbb{S}^1 \subset \mathbb{S}^3$ at the point $(e^{i \theta},e^{i\psi})$ is the vector $$\eta(\theta,\psi) := \frac{(e^{i \theta},-e^{i\psi})}{\sqrt{2}} \, .$$ Indeed, the scalar dot product of two vectors $\mathbb{v} = (v_1,v_2) ,\mathbb{w} = (w_1,w_2) $ in $\mathbb{C}^2$ is given by $$\mathbb{v} \cdot \mathbb{w} = real(v_1 \bar{w_1} + v_2.\bar{w_2}) \,. $$ To see that $\eta(\theta,\psi)$ do the job just notice that the normal space of $\mathbb{S}^1 \times \mathbb{S}^1 \subset \mathbb{C} \times \mathbb{C} =\mathbb{C}^2$ at the point $(\theta,\psi)$ is spaned by the two vectors $$(e^{i\theta},0) \, \, , \, \, (0,e^{i\psi})$$

so you just need to look for a linear combination of this two perpendicular to the normal of the sphere which is the position vector $\mathbf{p} = (e^{i \theta},e^{i\psi})$.

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  • $\begingroup$ Ofcourse, I get confused in finding the normal space, unused to complex scalar products or something. Thank you very much, now the result follows trivially $\endgroup$ – harajm Jun 27 '15 at 12:30
  • $\begingroup$ you are welcome. $\endgroup$ – Holonomia Jun 27 '15 at 12:58

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