Suppose $X$ and $Y$ are two smooth manifolds, with $\dim(X) = m < \dim(Y) = n$ is the inclusion map an immersion?

Question

Suppose $X \subseteq \mathbb{R}^k$ and $Y \subseteq \mathbb{R}^l$ are two smooth manifolds, with $\dim(X) = m < \dim(Y) = n$ is the inclusion map an immersion (where $X$ is a submanifold of $Y$)?

My Attempted Proof:

The inclusion map $i : X \to Y$ defined by $i(x) = x$, has derivative at a point $a \in X$ of $di_a(x) = 1$, where $1$ is the identity matrix $\left[\frac{I_m}{0}\right]$ and maps $T_a(X)$ into $T_a(Y)$, and is thus injective, because $T_a(X) \cong \mathbb{R}^m$ and for any vector $x \in T_a(X)$ we have $$\left[\frac{I_m}{0}\right][x] = \langle x_1, ...x_m, 0, ...0 \rangle \in T_a(Y) \ \ \ \ \ \square$$

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Is my proof correct? Is it rigorous enough?

Answer 1

There are some issues here. First of all, it is probably true that $X\subset Y$, right? The inclusion map $X\hookrightarrow Y$ assumes that $X\subset Y$. Indeed, your proof is accurate assuming that you can prove that the differential has the form $[\frac{I_m}{0}]$ you claim. This might not be immediately obvious though. You can rigorously show this by examining what a submanifold of $Y$ looks like in local coordinates about $a$.

Hint: A submanifold $X\subset Y$ of codimension $k$ is locally defined by the vanishing of $k$ coordinates for $Y$.

Answer 2

You have to first ask, does this map even make sense? Given $X,Y$ are two different manifolds what does $\iota(x) = x$ even mean? When we express a function in coordinate notation like this then a map $f: X \to Y$ is identified with its coordinate representation $\psi \circ f \circ \phi^{-1}$ (read below). Hence, if the dimensions of $X,Y$ aren't the same, how can we expect a coordinate representation to reflect that?

$\textbf{Suggestions}$: The only way for your problem to make sense is to maybe say $\iota(x) = \pi \circ x$ where $\pi$ is the projection onto the first $m$-coordinates. Otherwise, I think you really want to ask, given these conditions on the dimension, is every smooth map from $X$ to $Y$ necessarily an immersion? For this you will need the constant rank theorem.

$\bullet$ Constant rank theorem: Suppose $f: X \to Y$ is a smooth map between manifolds such that $Df(x)$ has rank $k$. Then there exists local coordinates $(U, \phi)=(U,x^1,...,x^{\textbf{dim}(X):=m})$ and $(V, \psi)=(V,y^1,...,y^{\textbf{dim}(Y):=n})$ such that $\psi \circ f \circ \phi^{-1}(x^1,...,x^m) = (x^1,...,x^k,0,....,0)$.

$\textbf{Recall}$: that $\psi \circ f \circ \phi^{-1}$ is just the coordinate representation of the map $f: U \subset X \to V \subset Y$ i.e we can safely write $f(x^1,...,x^m) = (x^1,...,x^k,0,....,0)$. Hence, it is because of this theorem that we have a canonical representation for the inclusion map.

$\textbf{Problem}$: With this theorem we can now pose problems that I think you were building your solution toward.

$1)$ if $\iota: X \to Y$ is a smooth map with constant rank and the dimension of $X$ is strictly less than $Y$ then $\iota$ is an immersion.

$2$) if $\iota: X \to Y$ is a smooth map such that $ \psi \circ \iota \circ \phi^{-1}(x^1,...,x^m) = (x^1,...x^m,0,...,0)$ for any charts $\phi, \psi$ on $X,Y$ respectively then $\iota$ is an immersion.

Answer 3

Yes, the inclusion $\imath:X^k\hookrightarrow Y^l$ ($k\le l$) is an immersion. To see this, note that for any $x$ in $X$, the map $D\imath_x:TX_x \to TY_x$ is the inclusion of tangent spaces for $\rm Id$ is a smooth extension of $\imath$ to an open neighborhood of $x$, and $D\imath_x$ is the restriction of $D({\rm Id})_{x} = \rm Id$. Hence the derivative $D\imath_x$ is injective, so $\imath$ is an immersion.