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Since $RP^2$ is compact and connected, its continuous image in $R$ is a closed interval. Let $f$ be this map. Suppose $RP^1 = f^{-1}(c)$. If $c$ is in the interior of $[a,b]$ then $RP^2$ \ $RP^1$ is not connected. Hence $c=a$ or $c=b$. WLOG assume $c=a$. I next want to argue that $f$ achieving a minimum on $RP^1$ implies $df = 0$ on $RP^1$, hence not surjective.As this problem arose in the context of a differential topology problem, I am hesitant to conclude that all partial derivatives of $f$ are zero at $p$, hence $df_p$ is null. I feel that any reference to partial derivatives requires mention of a chart. Therefore, how, in the language of differential topology, would I rigorously phrase such a conclusion?

Since first posting this question, I have some new thoughts. As I said in a later comment, this is as the question appeared on a qualifying exam. I think $f$ has to be smooth as opposed to merely $C^1$ since the Regular Level Set Theorem can only be applied with a smooth function. If $f$ is smooth, then the tangent plane of $R$ at $c$ can be spanned by $\frac{\partial f}{ \partial x_i} \Big|_p$ and if $c$ is the minimal value $f$ achieves in $R$, then given any $v \in T_p RP^2$, curve $\gamma: [0,1] \mapsto RP^2$ such that $\gamma(0) = p$, and $\gamma'(0) = v$, then $\sum_i v_i \frac{\partial f}{ \partial x_i} \Big|_p = df_p(v) = (f \circ \gamma)'(0) = 0$, hence $\frac{\partial f}{ \partial x_i} \Big|_p = 0$ for every $i$?

TerryL
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  • I edited your post to $\LaTeX$ify it. Remember your "$" signs! Also, are you sure you mean $c \in \Bbb R$, and not in the domain of $f$? – Robert Lewis Sep 16 '17 at 23:47
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    Suppose $f:[-1,1]\to \mathbb R$ maps $x\mapsto |x|$. It is continuous. It takes its minimum value at $0$, but $df_0$ does not exist. – kimchi lover Sep 16 '17 at 23:51
  • I have edited my question accordingly. Thanks for the suggestions! – TerryL Sep 17 '17 at 01:12
  • Hint: express $df_p$ in coordinates. – Vim Sep 17 '17 at 01:17
  • The connected, compact manifold in question is actually $RP^2$, so WLOG assume $p$ is in $U_3 := { [x_1, x_2 , x_3] : x_3 \neq 0}$ , which, with $\phi_3 ([x_1, x_2, x_3])= (x_1 / x_3, x_2/x_3 )$, $\phi_3^{-1} = [x_1, x_2, 1]$, defines a chart on $RP^2$. The image of f in $R$ is some $[c,d]$ and take as a chart any open set in $R$ containing $[c,d]$ and the identity map on $R$. So in coordinates, $df_p = [ df/dx_1 df/dx_2]$? I am not comfortable with such translation, hence the post. – TerryL Sep 17 '17 at 02:48
  • Furthermore, what conclusions does $f(p) = c$ allow me to make? I am hoping to get the conclusion is $df_p$ is zero map. – TerryL Sep 17 '17 at 02:51
  • The Regular Level Set Theorem (if it is what I think you are referring to) applies to $C^1$ maps; in general fibers will be as good as the map. – Alp Uzman Feb 11 '21 at 15:19

2 Answers2

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Let your connected compact space be $S^1=\{(x,y)\,|\,x^2+y^2=1\}\subset{\mathbb R}^2$. The function $f(x,y):=x$ assumes its minimum on $S^1$ at the point $p:=(-1,0)$, but $\nabla f(x,y)\equiv(1,0)$ is $\ne{\bf 0}$ at $p$.

Your idea of a "compact connected space" has to be made more precise.

Christian Blatter
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  • I did not include the actual question I was trying to solve as I thought I had isolated the remaining issue. Here it is exactly as it appeared on a qualifying exam at my university: "Regarding $S^1$ as the equator of $S^2$, we obtain $RP^1$ as a submanifold of $RP^2$. Show that $RP^1$ is not a regular level surface of any $C^1$ map from $RP^2$ into $R$." – TerryL Sep 17 '17 at 19:50
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If a submanifold $N\subset M$ is the preimage of a regular value for some smooth mapping $U\to\Bbb R^k$ ($N\subset U\subset M$, $U$ open), then the normal bundle of $N$ is $M$ must be trivial. But you can easily check that the normal bundle of $\Bbb RP^1$ in $\Bbb RP^2$ is not a trivial line bundle. (You can check this either analytically or topologically.)

Let me emphasize you should not be asking about a function defined on all of $\Bbb RP^2$ to start with. :)

Ted Shifrin
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  • Thanks for your comments! As I said in the comments to Christian Blatter's comments, the original question, as given on a past qualifying exam, is about a $C^1$ function defined on $RP^2$. Also, the differential topology part of the sequence was taught with Lee's $Introduction$ $to$ $Smooth$ $Manifolds$ as the textbook, but we didn't study the normal line bundle. Is there some way to approach the problem without this viewpoint? – TerryL Sep 17 '17 at 22:22
  • Do you understand how $\Bbb RP^1$ sits inside $\Bbb RP^2$ topologically? And what would the function have to look like around $\Bbb RP^1$ if the $\Bbb RP^1$ is the preimage of $0$ and $0$ is a regular value? – Ted Shifrin Sep 17 '17 at 22:30
  • Viewing $RP^2$ as $D^2/( x$ ~ $-x)$, $RP^1$ would be the boundary. This is why I thought that under any continuous map on $RP^2$ into the reals, the image of $RP^1$ would have to be one of the endpoints of a closed interval. Since the differential is surjective at every point in a regular value and the target space is $R$, then the image of the differential would have to be one dimensional. I am trying to get the conclusion that this is not the case by showing it is zero but I do not how to do this rigorously via the language of differential topology. – TerryL Sep 17 '17 at 22:58
  • Well, you're right. But what's your proof? And why does this contradict regular value? – Ted Shifrin Sep 17 '17 at 23:01