6
$\begingroup$

Let $X$ and $Y$ be smooth manifolds, let $i:X\to Y$ be the inclusion map, prove $di_x$ is the inclusion map from $T_x(X)$ to $T_x(Y)$. I know this is pretty basic, but can someone show me how to do this? The definition of derivatives on manifold are messing with me.

Thanks

$\endgroup$
  • $\begingroup$ What's your definition of tangent space and derivative? $\endgroup$ – xyzzyz Sep 30 '13 at 4:18
  • $\begingroup$ The definition is the image of the derivative of a parametrization of the manifold. $\endgroup$ – Pax Kivimae Sep 30 '13 at 4:26
  • $\begingroup$ I created paremtrizations that coincide on the shared subsets, and computed the derivative using that, but I am left with something I cannot process. $\endgroup$ – Pax Kivimae Sep 30 '13 at 6:43
  • $\begingroup$ I wasn't able to find this question anywhere online before and never bothered to ask about it on stackexchange, so I just assumed my proof of it was right. Good question, and I'm glad I found this! $\endgroup$ – Selene Auckland Jun 20 at 13:59
6
$\begingroup$

Let me provide some details regarding what I think the setup of your problem is, and hopefully that will clarify some things for you. If $i : X \to Y$ is an inclusion map, that implies that $X$ is a submanifold of $Y$. So suppose dim $Y = n$ and dim $X = k \leq n$. There are two ways to view $X,$ either 1) as a subset of $Y$ or 2) as an abstract $k$ dimensional manifold. These two viewpoints are related via the inclusion map $i : X \to X \subset Y$. Since $X$ is a submanifold of $Y$, around any point $x \in X$ we may choose a diffeomorphism $\phi :U \to \mathbb R^n$, where $U \ni x$ is an open set in $Y$, with the property that the first $k$ factors of $\phi$ map $\mathbb R^k$ onto $U \cap X$, i.e. we may arrange $\phi$ so that $$(x_1, \ldots, x_k, 0,\ldots, 0)$$ is the coordinate representation of $U \cap X$, where $x_i$ ($1\leq i \leq k$) are the first $k$ coordinate functions of $\phi$. In particular the restriction of $\phi$ to the first $k$ factors provides a coordinate chart for $X$ as an abstract manifold. Thus the coordinate representation of $i$ is nothing but the map $$(x_1, \ldots, x_k) \mapsto (x_1, \ldots, x_k, 0, \ldots, 0).$$ Now take the derivative of this map at any point $x_0$ in the coordinate chart and note that at the point $x_0$ you may identify the tangent space of $X$ with $\mathbb R^k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.