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Let $\phi :H\rightarrow M$ be a embedded submanifold of $M$. If I want to construct a smooth projection map $tan:\mathfrak X(M) \rightarrow \mathfrak X(H) $, one way is to use a Riemannian metric (whose existence is well known) to construct one (see [1] below). I want to understand this in a different way if possible.

Pointwise, I see the following happening:

At any point $p \in H$, consider $T_{\phi(p)}M$. Now, $\phi'(T_pH)$ is a linear subspace of $T_{\phi(p)}M$. By injectivity of $\phi'$, it is clear that $T_pH\cong \phi'(T_pH) $. Further, by standard linear algebra arguments and the definition of normal bundle, and above, one can easily say that: $$T_{\phi(p)}M \cong T_pH\oplus N_pH$$

Use the standard projectors $pr_i$ of the direct sum to construct the projector of the vector field as $$(tan(X))(p)=(\phi')^{-1}\circ pr_1(X(\phi(p)))$$ Ofcourse, the key caveat here is that we want the map to be smooth [2], and the problematic part is the $pr_1$ above. Now, I do not expect this map to be smooth, but I can't think of a counterexample to show that the above map is not smooth either.

[1] I know that given a Riemannian structure $g$ on $M$, one can induce one on $H$ as $h:=\phi^*g$. This yields a very natural smooth projection map as $tan:=h^\sharp \circ \phi^* \circ g^\flat$. But I am not interested in this.

[2] By smoothness of tan, I mean for all $X \in \mathfrak X(M)$, $tan(X) \in \mathfrak X(H)$ is smooth as a vector field of $H$.

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  • $\begingroup$ What exactly is your question? Whether your map $\tan$ is smooth or whether a Riemannian structure is necessary to construct projectors? $\endgroup$ Mar 2, 2019 at 11:50
  • $\begingroup$ Essentially both. If you can show that the map tan is smooth, then it basically means that a Riemannian structure is not really necessary to define one. So, I'd say the main question I want to know is if the map is smooth? If not, can you give me a counterexample? $\endgroup$
    – Sandesh Jr
    Mar 2, 2019 at 11:53
  • $\begingroup$ $\tan$ is a map between (more or less) function spaces. Are you really asking whether this map is smooth (with respect to some topology) or do you mean to ask whether for some $X \in \mathfrak{M}$, $\tan(X)$ is a smooth vector field on $H$? $\endgroup$ Mar 2, 2019 at 12:46
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    $\begingroup$ @SandeshJr: It's worth pointing out that it's not a standard linear algebra argument which guarantees that $\phi^{*}(TM) \cong TH \oplus NH$. The fact that there is a pointwise isomorphism is linear algebra but what you need for your projector to be smooth is a smooth decomposition. This is equivalent to choosing for each $p \in H$ a complement for $T_pH$ (or, more precisely, $d\phi|_p(T_pH)$) in $T_{\phi(p)} M$ in a smooth way. This is true because any short exact sequence of vector bundles splits but the proof itself usually uses a Riemannian metric so you don't really escape the metric. $\endgroup$
    – levap
    Mar 3, 2019 at 7:23
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    $\begingroup$ @SandeshJr: If you choose for each $p \in H$ an arbitrary isomorphism between $T_{\phi(p)}M$ and $T_p H \oplus N_pH$ then your projector won't be smooth. $\endgroup$
    – levap
    Mar 3, 2019 at 7:25

1 Answer 1

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I think you map $\tan$ is smooth. This can be checked locally, so let $p$ be an arbitrary point of $H$ and $U$ a sufficiently small open neighborhood of $p$. Take $V_1, \dots, V_n$ to be a local orthonormal frame of $T(U \cap H)$ (you can always obtain such a frame by starting with an arbitrary frame of $T(U \cap H)$ and applying Gram-Schmidt orthogonalization to it - locally it works for vector fields, too). First, the vector fields of this frame can be extended to orthogonal vector fields on $U$ (since $H \cap U$ is a embedded submanifold of $U$), and these can be extended to a orthogonal frame $V_1, \dots, V_m$ on $U$.

Now every vector field $X$ on $U$ can be expressed as a linear combination of $V_1, \dots V_m$ with smooth coefficients: $$ X = \sum_{k=1}^m f_k V_k $$ and it's easy to see that $$ \tan(X) = \sum_{k=1}^n f_k V_k $$ Since the $f_k$ are smooth, $\tan(X)$ is smooth.

Note that this also works for immersed manifolds.

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  • $\begingroup$ Yes, it seems correct to me. Using trivialisations also points to the same point that the map is smooth. It was surprising to me cause I thought that the orthogonal projectors were somehow the only way to project vector fields onto submanifolds. $\endgroup$
    – Sandesh Jr
    Mar 2, 2019 at 16:28
  • $\begingroup$ Reading @levap's comments above, I have realised that there is an inherent mistake above. In your step, by writing $tan(X)$ in the form, you have already made a specific choice for $pr_1$. It seems that in my definition, I have already introduced an inconsistency as there is no unique choice of $pr_1$. If nobody else writes an answer, I will write down one myself later. $\endgroup$
    – Sandesh Jr
    Mar 3, 2019 at 11:11
  • $\begingroup$ It don't understand you concern. $\operatorname{pr}_1$ is well defined at every $p \in H$ as the orthogonal projection onto $T_p H \subset T_p M$. This is obviously unique (after choosing a metric on $M$, of course). $\endgroup$ Mar 3, 2019 at 14:55
  • $\begingroup$ My original point was that one could do it without the orthogonal projection (which needs a metric). And then I realised that I was being stupid cause the projection map is only unique when one chooses a metric, and the tan that I defined in [2] pointwise turns out to be the same as the orthogonal projection. The way I defined $pr_1$ is through direct sum projection; the orthogonal projection is just a special case of it. That $pr_1$ from a direct sum need not be unique at all. $\endgroup$
    – Sandesh Jr
    Mar 3, 2019 at 18:54

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