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Consider a smooth manifold:

$$M=\{(x,y,z)\in\mathbb R^3 : F(x,y,z)=0\}$$

where $F(x,y,z)=x^2+y^2+z^2-1$.

A tangent space at point $p=(x_0,y_0,z_0)=(1,0,0)$ can be defined as

$$T_pM=\text{Ker}(dF(p))=\left\{(x,y,z)\in\mathbb R^3 : \begin{pmatrix} 2x_0 & 2y_0 & 2z_0 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0\right\}=\{(x,y,z)\in\mathbb R^3 : x=0\}\tag{1}\label{eq:1}$$

where $dF:\mathbb R^3\to\mathbb R$ is a Jacobian of $F$.

But in many differential geometry textbooks I skimmed, it seems like tangent spaces are defined either as a set of derivations or set of equivalence classes of smooth curves. In terms of derivations, tangent space above is now defined as

$$T_p M=\text{span}\left\{\left.\frac{\partial}{\partial x}\right|_p, \left.\frac{\partial}{\partial y}\right|_p\right\}\tag{2}\label{eq:2}$$

I cannot see how \eqref{eq:1} and \eqref{eq:2} are related to each other. I wonder if \eqref{eq:1} and \eqref{eq:2} equivalent in sense that there is an isomorphism between them. If so, what would an isomorphism look like?

Jimmy Yang
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  • You need a chart $(x,y)$ on an open subset. To make this as unconfusing as possible, use projection of the upper (lower) hemisphere onto the unit ball in the $(x,y)$-plane. – Ted Shifrin Jan 09 '25 at 00:24
  • @TedShifrin I just realized that I forgot $\left.\frac{\partial}{\partial z}\right|_p$ in my second definition of tangent space. Can you elaborate how projection is relevant here? – Jimmy Yang Jan 09 '25 at 01:01
  • No, you’re wrong to add that. This is a $2$-dimensional manifold. Review your definitions carefully and perhaps don’t use the same letters. – Ted Shifrin Jan 09 '25 at 01:10

1 Answers1

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(2) isn’t the right definition. It only makes sense if $(x,y)$ is a local coordinate system for $M$, in particular, they’re not necessarily the same $x,y$ appearing in $F(x,y,z)=0$. Where exactly did you see (2)?

Anyway, I know of four equivalent definitions of submanifolds of $\Bbb{R}^n$; see this question and my answer for the equivalence. Corresponding to these four definitions of submanifolds, one has corresponding descriptions for the tangent spaces.

Let’s start with a somewhat general but intuitive definition:

Definition.

Let $M$ be a $k$-dimensional embedded submanifold of $\Bbb{R}^n$ and $p\in M$. We define $T_pM$ to be the subset of $\Bbb{R}^n$ consisting of the set of all velocity vectors $\gamma’(0)$ of smooth curves $\gamma:(-\epsilon,\epsilon)\to \Bbb{R}^n$ such that $\text{image}(\gamma)\subset M$ and $\gamma(0)=p$.

In the more general case of abstract manifolds, we can’t speak of $\gamma’(0)$ directly as a limit of difference quotients, rather we need to work with equivalence classes of smooth curves. Anyway, this is just one extra layer of definitions, and won’t affect what the core of what I’m about to say below.

Theorem (Various Descriptions of Tangent Spaces).

Let $M$ be a $k$-dimensional embedded submanifold of $\Bbb{R}^n$ and let $p\in M$. Fix the following notation: suppose $U\subset \Bbb{R}^n$ is an open neighbourhood of $p$ such that we have the following:

  • Graph description of $M$: there is an open subset $A\subset\Bbb{R}^k$ and a smooth function $f:A\to\Bbb{R}^{n-k}$ such that $M\cap U=\text{graph}(f)$ (super strictly speaking I should assume there exists a permutation of the coordinates, $\sigma$, such that $\sigma[M\cap U]=\text{graph}(f)$). Suppose $a\in A$ is the point such that $p=(a,f(a))$.

  • Local level set: there is a smooth submersion $F:U\to\Bbb{R}^{n-k}$ and a $c\in\Bbb{R}^{n-k}$ such that $M\cap U=F^{-1}(\{c\})$.

  • Local slice chart: there is an open set $V\subset\Bbb{R}^n$ and a diffeomorphism $\Phi:U\to V$ such that $M\cap U=\Phi^{-1}[V\cap (\Bbb{R}^k\times\{0_{\Bbb{R}^{n-k}}\})]$.

  • Local parametrization: there is an open set $W\subset\Bbb{R}^k$ and a smooth map $\alpha:W\to\Bbb{R}^n$ such that $\text{image}(\alpha)=M\cap U$, $\alpha$ is an injective immersion, and $\alpha^{-1}:M\cap U\to W$ is continuous.

With notation as above, we have the following equalities for the tangent space $T_pM$: \begin{align} T_pM=\text{graph}(Df_a)=\ker(DF_p)=(D\Phi_p)^{-1}(\Bbb{R}^k\times\{0\})=\text{image}(D\alpha_{\alpha^{-1}(p)}). \end{align} In particular, $T_pM$ is a $k$-dimensional subspace of $\Bbb{R}^n$.

All these equalities should be intuitive:

  • $T_pM=\text{graph}(Df_a)$ is just saying that since $M$ is locally a graph, it follows that the tangent space (i.e linear approximation to $M$ at $p$) is the graph of the linear approximation.
  • $T_pM=\ker(DF_p)$ just says that since $M$ is locally a level set of $F$, the tangent space is the zero-level set of the linear approximation of $F$.
  • $T_pM= (D\Phi_p)^{-1}(\Bbb{R}^k\times \{0\})$ just says that since $M$ is locally the preimage of an open subset of $\Bbb{R}^k$, the tangent space is the preimage of $\Bbb{R}^k$ under the linear approximation.
  • $T_pM=\text{image}(D\alpha_{\alpha^{-1}(p)})$ just says that since $M$ is locally the image of $\alpha$, the tangent space is the image of the linear approximation of $\alpha$ (at the respective point $\alpha^{-1}(p)$).

Now, let’s outline the proof of the various equalities.

  • the fact that $T_pM=\text{graph}(Df_a)$ is the easiest of all, and I leave the details to you. You simply have to consider curves of the form $t\mapsto (a+tv, f(a+tv))$ in the graph. This also shows that $\dim T_pM$ equals the dimension of the domain of $Df_a$, namely the dimension of $\Bbb{R}^k$, which is $k$.
  • To show $T_pM=\ker(DF_p)$, observe that for any smooth curve $\gamma:(-\epsilon,\epsilon)\to M$ through $p$, if we shrink $\epsilon$ small enough then the image of $\gamma$ lies in $M\cap U$, and so $F\circ \gamma$ is constantly equal to $c$. Thus, $DF_p(\gamma’(0))=0$, showing that $\gamma’(0)\in \ker(DF_p)$, and thus $T_pM\subset\ker(DF_p)$. The former is a $k$-dimensional subspace as I mentioned above, while the latter is $k$-dimensional by the rank-nullity theorem and the fact that $DF_p$ is surjective (by hypothesis).
  • To show $T_pM=(D\Phi_p)^{-1}(\Bbb{R}^k\times \{0\})$, just take any velocity vector $\gamma’(0)\in T_pM$. Then, by shrinking the domain $(-\epsilon,\epsilon)$ sufficiently, we again have that $\gamma$ has image in $M\cap U$, and thus $\Phi\circ\gamma$ has image in $V\cap (\Bbb{R}^k\times\{0\})$, and thus $(\Phi\circ\gamma)’(0)=D\Phi_p(\gamma’(0))\in\Bbb{R}^k\times\{0\}$ (since $(\Phi\circ\gamma)’(0)$ is a limit of difference quotients and $\Bbb{R}^k\times\{0\}$ is a closed subset of $\Bbb{R}^n$). This shows $T_pM\subset (D\phi_p)^{-1}(\Bbb{R}^k\times\{0\})$; finally since $D\Phi_p$ is an isomorphism, tbe latter is a $k$-dimensional space, so we have equality.
  • Lastly, to show $T_pM=\text{image}(D\alpha_{\alpha^{-1}(p)})$, observe that for each $w\in\Bbb{R}^k$, by openness of $W$, there is an $\epsilon>0$ such that for all $|t|<\epsilon$, we have $\alpha^{-1}(p)+tw\in W$. So, $\gamma:(-\epsilon,\epsilon)\to \Bbb{R}^n$ defined as $\gamma(t):=\alpha(\alpha^{-1}(p)+tw)$ is a smooth curve in $M\cap U$ through the point $p$. Next, we have $\gamma’(0)=D\alpha_{\alpha^{-1}(p)}(w)$ by the chain rule, and so we see that $\text{image}(D\alpha_{\alpha^{-1}(p)})\subset T_pM$. Since $\alpha$ is an immersion, it means $D\alpha_{\alpha^{-1}(p)}:\Bbb{R}^k\to\Bbb{R}^n$ is an injective linear map, so the image has dimension $k$, and thus we have equality of the two spaces.

This proves the equality of all the descriptions of the tangent spaces to submanifolds.


Local Coordinate Description.

When talking about local coordinates for submanifolds, we’re really talking about the third definition, i.e what I called the local slice chart definition. Observe that because $\{e_1,\dots, e_k\}$ is a basis for $\Bbb{R}^k\times\{0\}\subset \Bbb{R}^n$ and $D\Phi_p$ is an isomorphism, it follows that $\{(D\Phi_p)^{-1}(e_i)\}_{i=1}^k$ forms a basis for $T_pM$.

For each $i\in\{1,\dots, k\}$, let us define $\xi^i:= \Phi^i|_{M\cap U}$, i.e the function $\xi^i$ is the $i^{th}$ component function of $\Phi:U\to V\subset\Bbb{R}^n$, restricted to $M\cap U$. Then, the collection $(M\cap U, (\xi^1,\dots,\xi^k))$ is said to comprise a local coordinate system for $M$ around $p$. If you know about abstract manifolds, you can easily verify that the collection of all such comprise a smooth atlas for $M$ (look up any good book for more details, e.g Lee’s book on smooth manifolds). Now, one introduces the following notation: \begin{align} \frac{\partial}{\partial \xi^i}(p):= (D\Phi_p)^{-1}(e_i)\in T_pM.\tag{$*$} \end{align} So, by my previous paragraph, one trivially has \begin{align} T_pM=\text{span}\left\{\frac{\partial}{\partial \xi^i}(p)\right\}_{i=1}^k.\tag{$**$} \end{align}

Notice that the equality $(**)$ doesn’t have much content; it’s a simple consequence of the notation $(*)$ and the previously established fact that $T_pM=D\Phi_p^{-1}(\Bbb{R}^k\times\{0\})$. So, the real question is why one even introduces the notation $(*)$. For that you can read up Lee’s book or search the site because it’s been asked several times.

Finally, I should mention that you should observe very carefully that the functions $\xi^i$ are not the same as the coordinate functions $x^i$ on $\Bbb{R}^n$. It is probably your double usage of $x^i$ (or $(x,y)$) for both the coordinate functions on the ambient space and the submanifold which is confusing you. Bottom line is you should review the definitions carefully.

peek-a-boo
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  • Thank you very much for the encyclopedic answer! Can you define what is $D$? It seems like all instances of $0$ are $n-k$ dimensional column vector of zeros, i.e. $0_{\mathbb{R}^{n-k}}$ right? – Jimmy Yang Jan 09 '25 at 01:59
  • How $T_p M$ defined in $(**)$ is a $k$-dimensional subspace of $\mathbb R^n$? I thought it was a subset of set of all functions from $C^\infty(\mathbb R^n)$ to $\mathbb R$. – Jimmy Yang Jan 09 '25 at 02:03
  • @JohnDavies $D$ is the Frechet derivative for maps between open subsets of Banach spaces (in this case finite-dimensional ones)… or in less scary terms the “total derivative”. And yes you’re right about the $0$ in ${0}$. And regarding $()$ you’re asking the wrong question. I already said that $()$ is trivially true, and that $(*)$ is where the notation comes from. So if you have questions about that then that’s what you should ask (but it’s been asked several times on the site already so look at those first). – peek-a-boo Jan 09 '25 at 02:23
  • All throughout my answer, I started with only one definition of $T_pM$, and it was as a subset of $\Bbb{R}^n$, they I proved it was equal to a bunch of other stuff. In particular, $(**)$ is not a definition. – peek-a-boo Jan 09 '25 at 02:24
  • Finally, I have tons of answers about relating abstract definitions of tangent spaces to actual subspaces of $\Bbb{R}^n$ when we deal with submanifolds. See this and all the sublinks. – peek-a-boo Jan 09 '25 at 02:35
  • I still don't get it. So, $\frac{\partial}{\partial \xi^i}(p)$ is an element of $\mathbb R^n$, right? But I read first few pages of chapter 3 of Lee's book, I still got the impression that $\frac{\partial}{\partial \xi^i}(p)$ (in Lee's notation $\left.\frac{\partial}{\partial \xi^i}\right|_p$) is treated as a map from $C^\infty(\mathbb R^n)$ to $\mathbb R$. – Jimmy Yang Jan 09 '25 at 03:11
  • Read the link about relating abstract definitions to concrete ones. – peek-a-boo Jan 09 '25 at 03:14
  • I checked out the links you provided but I am not sure if I got it. So in summary, $T_p M$ you "defined" in $(**)$ is a subset of $\mathbb R^n$. But some authors like Lee define $T_p M$ in such way that it is a subset of the set of all functions $f:C^\infty(\mathbb R^n)\to\mathbb R$. In this case, the image of $T_p M$ under the linear isomorphism $\Phi:T_p M\to\mathbb R^n$ is a subset of $\mathbb R^n$... – Jimmy Yang Jan 09 '25 at 06:17
  • Are all implementations of tangent spaces isomorphic to each other canonically? – Jimmy Yang Feb 11 '25 at 18:27
  • @weeab00 the way I described things here, they are all equal. – peek-a-boo Feb 11 '25 at 18:30
  • I know. I am just wondering if all tangent space definitions are canonically isomorphic to each other, just like how a vector space is canonically isomorphic to its double dual space. For example, is your definition of tangent space is canonically isomorphic to tangent space defined as a set of derivations or as a set of equivalence classes of smooth curves? I already read that isomorphism exists but I am not sure if they are canonical. – Jimmy Yang Feb 11 '25 at 18:38
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    @weeab00 yup it’s easy to write down some of the maps (though perhaps to prove bijectivity you may have to go down to a chart, atleast to verify dimensions, analogous to how for the double duality map in finite dimensions, showing it’s an isomorphism relies on having a basis/invoking rank-nullity). For example, given an equivalence class of curves $\Gamma=[\gamma]$, one can define a derivation $X_{\Gamma}: f\mapsto (f\circ\gamma)’(0)$. You can check this is well-defined by the definition of the equivalence relation and by the chain rule of analysis, and investigate its in/sur-jectiity etc. – peek-a-boo Feb 11 '25 at 19:14