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Define $F: \mathbb{R}^3 \rightarrow \mathbb{R}^4$ by $F(x,y,z) = (x^2-y^2,xy,xz,yz)$. Notice that $F(x,y,z) = F(-x,-y,-z)$ so that $F|_{S^2}$ defines a mapping $\tilde{F}: \mathbb{R}P^2 \rightarrow \mathbb{R}^4$. We want to show that $\tilde{F}$ is a smooth imbedding.

Proof:

We can show that $\tilde{F}$ is smooth and by above it is well defined. Then we also know that $\mathbb{R}P^2$ is compact. Next we can see that $\tilde{F}$ is 1-1 as well. I can prove all of this in detail it is not where my problem lies. Now we know that if $X$ is compact and $f:X \rightarrow Y$ a 1-1 immersion we have that it is a smooth imbedding. The problem is why is this an immersion. Our definition of immersion is that $f:X \rightarrow Y$ is an immersion if the rank of the jacobian matrix is equal to the dimension f $X$. In our case the dimension of $\mathbb{R}P^2$ is 2 so i need to show that the dimension of the jacobian matrix is 2. The jacobian matrix is given by: \begin{array}{lcr} 2x & -2y & 0 \\ y & x & 0 \\ z & 0 & x \\ 0 & z & y \end{array}

Now what is the rank of this matrix. My problem is we can look at the equialence class of the point (0,0,1) which lies on the sphere and gives me rank 2 at that point. We can also use the equivalence class of (1,0,0) which lies on the sphere, but then gives a rank of 3 at that point. To say its an immersion at every point we must have that the rank is constant, which it is not here. Also we must have that it must be rank 2 to use the thereom stated above. Any help would be appreciated I do not see what I am missing.

InsideOut
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Michael N
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1 Answers1

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Now $\tilde F : \mathbb{RP}^2 \to \mathbb R^4$, so the Jacobian should be a mapping

$$T_{\bar x} \tilde F : T_\bar{x} \mathbb{RP}^2 \to \mathbb R^4, $$

where $\bar x \in \mathbb {RP}^2$. Note that for each $\bar x = [x, y, z]$, we have the identicafication $$T_\bar{x} \mathbb{RP}^2 = \{ \bar a= (a, b, c) \in \mathbb R^3 : \bar a \cdot \bar x = 0\}.$$

Note that this can be considered as a subspace of $\mathbb R^3$. So if the rank $DF$ is $3$, then when restricted to this subspace $T_\bar{x} \mathbb{RP}^2$, it must be of rank $2$.

Actually you need to be a bit careful when $DF$ is of rank $2$. This happens only when $\bar{x} = [0,0,1]$ (check!). At this point, $$T_{\bar x} \mathbb{RP}^2 = \text{span} \langle (1, 0, 0), (0,1,0)\rangle $$

and so the image of $T_\bar{x}\tilde F : T_\bar{x} \mathbb{RP}^2 \to \mathbb R^4$ is $\text{span}\langle (0,0,1,0), (0,0,0,1)\rangle$, which is two dimensional. Thus $T_\bar{x} \tilde F$ is always injective and so it is an immersion.