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My professor gave us this exercise.

Let $S^2=\{(x_1,x_2,x_3)\in\mathbb{R}^3|(x_1)^2+(x_2)^2+(x_3)^2=1\}$ be the unit sphere in $\mathbb{R}^3$, and let's consider the function $f\colon S^2\to\mathbb{R}^6$ defined by

$$ f(x_1,x_2,x_3)=((x_1)^2,(x_2)^2,(x_3)^2,x_2x_3,x_1x_3,x_1x_2). $$

Prove that $f$ is an immersion but is not injective. Besides, write the equations for $f(S^2)\subset\mathbb{R}^6$ and prove that $f(S^2)$ is an embedded submanifold of $\mathbb{R}^6$.

Solution

For the first question, it is sufficient to write down the jacobian matrix $J$ of the function f and note that it is not possible for all the determinants of the $3\times3$ submatrices of $J$ to be zero simultaneously (provided that we are considering only the points of the 2-dim sphere), so the differential is everywhere injective on the sphere and we are done. Besides, $f$ is not injective because, for example, $f(x_1,x_2,x_3)=f(-x_1,-x_2,-x_3)$ ($(x_1,x_2,x_3)$ lies on the 2-dim sphere if and only if $(-x_1,-x_2,-x_3)$ does).

But I am just stuck with the second question, I don't get what my professor means with "write the equations for $f(S^2)$"... The only equation I see is satisfied by a point in $f(S^2)$ is $x_1+x_2+x_3=1$. I'm wondering, how many equations there should be for $f(S^2)$? And how can I find them? Only by means of algebric manipulations of the components of $f$, or is there some "sistematic" way to find them?

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  • $\begingroup$ A point $(x,y,z,u,v,w)$ is in the image if $x^2+y^2+z^2 = 1$ and $\frac{uv}{w} + \frac{vw}{u} + \frac{uw}{v} = 1$. $\endgroup$ – Zestylemonzi Sep 11 '16 at 12:24
  • $\begingroup$ Oops you need to be careful of the cases when $u,v$ or $w$ are $0$. This is easy to take care of though. $\endgroup$ – Zestylemonzi Sep 11 '16 at 12:47
  • $\begingroup$ Why $x^2+y^2+z^2=1$ and not $x+y+z=1$? A point in $f(S^2)$ is a point which belongs to $\mathbb{R}^6$ and can be written in the form prescribed by $f$, for some point that belongs to the 2-shpere, and so should satisfy $x+y+z=1$ I think. $\endgroup$ – Vladimir Sep 11 '16 at 13:30
  • $\begingroup$ Yes you're right! Sorry, I misread the equation. $\endgroup$ – Zestylemonzi Sep 11 '16 at 15:28
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Note $f(\mathbb S^2)$ is a closed set in $\mathbb R^6$ and thus there is a smooth function $F :\mathbb R^6 \to \mathbb R$ so that $F^{-1}(0) = f(\mathbb S^2)$. Of course this does not say anything about the geoemtry of $f(\mathbb S^2)$. So we do something else.

Write the coordinate of $\mathbb R^6$ as $(x, y, z, w, u, v)$. Since $f(\mathbb S^2)$ should be two dimension (it's immersed) and thus are cut out by four equations. By inspection, we have

$$\begin{split} x+y+ z&=1 \\ xy &= v^2 \\ yz &= w^2 \\ zx &= u^2 \end{split}$$

equation $(2)-(4)$ implies that $xy, yz, zx \ge 0$. Together with $(1)$ we have $x, y, z \ge 0$. Write $$(x, y,z) = (x_1^2, x_2^2, x_3^2)$$ for some $x_1, x_2, x_3 \in \mathbb R$. Then

$$\begin{split} v &= \pm x_1x_2 \\ w &= \pm x_2x_3 \\ u &= \pm x_3x_1.\end{split}$$

If we require $w, u, v \ge 0$, then we recover

$$(x_1^2 , x_2^2, x_3^2 , x_2x_3, x_1x_3, x_1x_2), \text{with }\ x_1^2 + x_2^2 + x_3^2 = 1. $$

Thus $f(\mathbb S^2)$ is the set cut out by the above equations in the quadrant $w, u, v\ge 0$.

Now we show that $f(\mathbb S^2)$ is embedded. Indeed it is clear that if $x, y\in \mathbb S^2$ and $f(x) = f(y)$, then $x = \pm y$ (check). Then $f$ descends to an injective immersion $\tilde f :\mathbb{RP}^2 \to \mathbb R^6$, which is embedded as $\mathbb {RP}^2$ is compact. Thus $f(\mathbb S^2)$ is an embedded submanifold in $\mathbb R^6$ homeomorphic to the real projective plane.

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I don't know about finding the equations describing the surface in $\mathbb R^6$, but I'll demonstrate how we might apply abstract machinery to answer the second question without explicit equations. This probably isn't the kind of answer you're looking for (i.e. may be unnecessarily abstract for the problem at hand), but I think it will be good for you to see the technique anyhow. Hopefully it helps you solve future problems.


You have already shown that $f(x_1,x_2,x_3) = f(-x_1, -x_2, -x_3)$, but we can also note that this is the only transformation of coordinates $(x_1,x_2,x_3)$ that preserves the value of $f$. Why? Well, for a start we know that we can only flip the signs of $x_1$, $x_2$ and $x_3$ because otherwise, one of first three coordinates $(x_1^2, x_2^2, x_3^2)$ would change. You can easily see from the remaining coordinates ($x_1 x_2$, $x_2 x_3$, $x_3 x_1$) that flipping the sign of one of these $x_i$ requires changing another, and then the final one. So inversion of $(x_1, x_2, x_3)$ is the only transformation under which $f$ is invariant. This tells us that in some sense, $f(S^2)$ is equivalent to $S^2/\sim$ where $\sim$ is the equivalence relation $\mathbf x \sim -\mathbf x$. So the remainder of the proof consists of two steps:

  1. Show that as topological spaces, $S^2/\sim$ is homeomorphic to $f(S^2)$.
  2. Show that $S^2/\sim$ is a manifold.

As an aside, this quotient structure is a very common way to represent the manifold $\mathbb {RP}^2$ called the real projective plane.

$S^2 / \sim$ is homeomorphic to $f(S^2)$

There is an inherited map $g : S^2 / \sim \rightarrow f(S^2)$ that maps $g : [\mathbf x] \mapsto \mathbf x \mapsto f(\mathbf x)$. I'll show that this is a homeomorphism.

I'll assume that $f$ is continuous and an open mapping (its inverse is continuous). By definition of the quotient topology of $S^2/\sim$, the canonical projection from an element $\mathbf x$ in $S^2$ to its equivalence class $[\mathbf x]$, the map denoted $\pi$, is continuous and an open mapping. This is because $U \subseteq S^2/\sim$ is defined to be open iff $\pi^{-1}(U)$ is open. Consequently, the map $g$ mapping sets in $S^2/\sim$ to sets in $f(S^2)$, given by $g = f \circ \pi^{-1}$, is composed of continuous, open mappings. This means it itself is continuous and an open mapping. Furthermore, $g$ is injective because we ensured that it would be so by quotienting out points of $S^2$ which would give duplicate values of $f$. $g$ is also surjective because we take its domain to be $f(S^2) = g(S^2/\sim)$. Hence $g$ is a homeomorphism.

$S^2/\sim$ is a manifold

We are looking to show that every $\mathbf y \in S^2 / \sim$ has an open neighbourhood homeomorphic to $\mathbb R^2$. First, let $\mathbf y = \{\mathbf x, -\mathbf x\}$. Since $S^2$ is a manifold, we can take a neigbourhood $U_{\mathbf x} \subset S^2$ that is homeomorphic to $\mathbb R^2$. This defines a companion neigbourhood $U_{- \mathbf x}$ on the opposite side of the sphere which is related by the inversion operation $\mathbf a \mapsto - \mathbf a$. Note that this operation is a homeomorphism. An open neighbourhood of $\mathbf y$ is $V_{\mathbf y} = \{\{\mathbf a, \mathbf - \mathbf a\} : \mathbf a \in U_{\mathbf x}\}$. Basically this is a set of pairs of poles of the sphere, where one of each pair lies in $U_{\mathbf x}$. The projection from $V_{\mathbf y}$ onto $U_{\mathbf x}$ by taking that one pole lying in $U_{\mathbf x}$ is a homeomorphism (I'll leave you to think about this). And so $V_{\mathbf y}$ may be mapped to $\mathbb R^2$ by a pair of composed homeomorphisms (first onto $U_{\mathbf x}$ and then onto $\mathbb R^2$) and so we have constructed a neighbourhood of $\mathbf y$ homeomorphic to $\mathbb R^2$.


This may sound needlessly complicated (maybe it is), but it has some far-reaching consequences. For instance, we now know that if we have an arbitrary continuous and open map between topological spaces $f : X \rightarrow Y$, then $f(X)$ is homeomorphic to $X/\sim$ where $\sim$ identifies elements of $X$ with the same value in $Y$. Consequently, to determine the topological structure of $\mathrm{range} (f)$, including whether or not it is a manifold, we may look instead at the topological structure $X / \sim$.

This formalises a notion which seems rather obvious when one thinks about it.

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