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Let $\iota : S \hookrightarrow M$ be an inclusion that serves as an injective immersion between real manifolds of dimension $k$ and $n$, respectively. Fix $p \in S$. Then we have a linear embedding $\iota_{*, p} : T_{p}S \hookrightarrow T_{p}M$, and we wish to compute $\iota_{*, p}(T_{p}S)$. It is easy to see that $\iota_{*, p} (T_{p}S) \leqslant \{X_{p} \in T_{p}M : X_{p}(f) = 0 \text{ for all } f \in C^{\infty}(M) \text{ such that } f|_{S} = 0\}$.

Moreover, if $S$ is a regular submanifold (or embedded submanifold), meaning that we can take a chart $(U, x^{1}, \cdots, x^{n})$ near $p$ in $M$ so that $(U \cap S, x^{1}, \cdots, x^{k})$ is a chart near $p$ in $S$, one can prove that the above inclusion becomes equality.

I would like to know some examples where this equality does not hold. I imagine there must be some evident examples, but I have not succeeded in finding them. I appreciate your help.

user123454321
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1 Answers1

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CORRECTED:

If $S$ fails to be embedded in a neighborhood of $p\in M$, equality can fail. Consider a figure 8 in $\Bbb R^2\subset\Bbb R^3$ that is the image of a 1-1 immersion. Then if the two branches are tangent to the $x$- and $y$- axes at $p=0$, your right-hand side will be the $xy$-plane in $\Bbb R^3$. So this is an example where the right-hand side is strictly larger than $\iota_{*p}(T_pS)$ but strictly smaller than $T_pM$.

Ted Shifrin
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  • Can I ask you more about how to compute the right-hand side? Even in your example, the right-hand side must be a linear subspace of $T_{0}\mathbb{R}^{2}$, so to me it seems like the only case we do not get the equality is when the right-hand side takes all the tangent space $T_{0}\mathbb{R}^{2}$. – user123454321 Jun 04 '14 at 06:10
  • Sorry for the delay. I didn't see that you'd responded. Yes, you're right. I screwed that up. Since $f$ is smooth on $\Bbb R^2$, if it has directional derivative $0$ in two linearly independent directions at $p$, it has to have derivative $0$ at $p$. But if $S$ has higher codimension in $M$, we'll get something lower-dimensional for the right-hand-side. – Ted Shifrin Jun 05 '14 at 03:13
  • No need to apologize; I am sorry that I need to ask one more question. Given $f|{S} = 0$, we need to show $\partial f/\partial x|{0} = 0 = \partial f/\partial y|{0}$. We have $\partial f/\partial x|{0} = \lim_{h \rightarrow 0}f(h, 0)/h$, so if we "deform" figure 8 a little so that it touches $x$-axis and $y$-axis more so that $(h, 0) \in S$ for sufficiently small $|h| > 0$ (and similarly to the $y$-direction), we are done (right?). But I do not know why partials $\partial f/\partial x|{0}, \partial f/\partial y|{0}$ are zero for just "ordinary" figure 8 that are tangent to the axes. – user123454321 Jun 06 '14 at 02:51
  • The chain rule tells you that the directional derivative of (a differentiable function) $f$ in direction $v$ at $p$ is $(f\circ \phi)'(0)$ for any curve $\phi$ with $\phi(0)=p$ and $\phi'(0)=v$. – Ted Shifrin Jun 06 '14 at 03:03
  • So you parametrize two basis elements by taking two such $\phi$'s and show they evaluate $f$ as $0$. That makes sense, and thanks much for your patience. – user123454321 Jun 06 '14 at 04:57