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I am trying to gain some intuition about tangent spaces. According to the definition here, a tangent vector is an equivalent class of curves on the manifolds. Does this mean that the tangent vector lie on the manifold?

This does not seem equivalent to the tangent plane we learn in $\mathbb{R}^n$. How do I derive the tangent plane in $\mathbb{R}^n$ from the general definition of the tangent spaces?

For example, I am interested in the surface $(x, y, f(x,y))$ for a smooth function $f$. I consider the chart $\gamma(x,y,f(x,y)) = (x,y) \in \mathbb{R}^2$. How do I define curves and the equivalence class of curves so that I end up with the tangent plane? Could you please help with this example so I can gain intuition? Thank you!

wut
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    No, it does not. Do you know what an equivalence relation and an equivalence class mean? – Moishe Kohan Dec 02 '24 at 02:38
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    The tangent vector is tangent to each of the curves in the equivalence class. It does not lie on the manifold; it just touches it at a point on the manifold where all the curves in the equivalence class meet. – Anuradha N. Dec 02 '24 at 02:43
  • @MoisheKohan I think I know but maybe I do not know. Does it not contain its element (curves) ? Could you please explain it for me? How can I describe such an equivalence class on the space/manifold? – wut Dec 02 '24 at 02:47
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    Take a look for instance, here for the discussion of equivalence relations. One has to learn this before trying to study differential topology. – Moishe Kohan Dec 02 '24 at 02:49
  • @AnuradhaN. then how do we define the tangent vector? I am very confused, according to the definition here they even say let $\gamma \in \gamma'(0)$, which means $\gamma'(0)$ contains a curve? – wut Dec 02 '24 at 02:52
  • @MoisheKohan I think I understand it correctly. So the tangent vector at a point $p \in M$ is a set of curves that are equivalent in the directional derivative sense? How do we match this definition to the case for a surface in $\mathbb{R}^n$? – wut Dec 02 '24 at 02:55
  • It is a theorem that there exists an isomorphism between the two vector spaces given by the different definitions (there is one, more common, definition using derivations). What textbook are you reading? – Moishe Kohan Dec 02 '24 at 03:00
  • Let $X$ be the surface in $\mathbb{R}^3$. Take a normal vector $N$ of the surface at point $p$. Let $T$ be the plane passing through $p$, and normal to $N$. Consider a straight line passing through the point $p$ in this plane. Near $p$, you can project the straight line to generate a curve in the surface $X$. Now you need to show that those curves give the full representatives of the equivalence classes. – J1U Dec 02 '24 at 03:03
  • @wut think of $\gamma'(0)$ as the tangent vector to each curve $\gamma$ in the equivalence class passing through the point 0 ($\gamma \in \gamma'(0)$;just symbolises this). – Anuradha N. Dec 02 '24 at 03:04
  • @MoisheKohan I do not have a specific book, just self-study so I am very confused. I also did not find any discussion about isomorphism in the link you shared. And can you link this definition of tangent space and tangent vector to the case of the surface in $\mathbb{R}^n$? – wut Dec 02 '24 at 03:04
  • Most likely, this was discussed several times on this site, take a look for insatnce here. But my suggestion is to use a textbook. There are many suggestions given on MSE, e.g. here. – Moishe Kohan Dec 02 '24 at 03:07
  • @MoisheKohan Do you mean that we can find an isomorphism from the quotient set to some surface in $\mathbb{R}^n$? – wut Dec 02 '24 at 03:10
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    It is not just a set but a vector space. I am talking about an isomorphism of vector spaces as discussed in a linear algebra class. Did you study linear algebra? You have to learn it before studying differential topology. – Moishe Kohan Dec 02 '24 at 03:12

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I'm gonna go ahead and try to talk about this all in $\mathbb R^2$. Let's look at a point $P = (3, 5)$, just to be concrete. A typical tangent vector at $P$ is something like $$ v = \pmatrix{2\\1}, $$ which we tend to think of as an arrow pointing about ENE, with the arrow starting at $P$. Some authors instead say that a tangent vector at $P$ is a pair like $(P, v)$, with the rule that the "$P$" is "just along for the ride", so that addition is defined by a rule like $$ (P, v) + (P, w) = (P, v+w) $$ where that last addition is just ordinary addition of vectors in $\mathbb R^2$.

Presumably you can see that this set of pairs is in a 1-1 correspondence with $\mathbb R^2$, with $u \leftrightarrow (P, u)$. So it doesn't really matter which representation you choose.

As a practical matter, one use of a vector like $v$ (or $(P, v)$) is to describe 'directional derivatives': if we have a function $$ f: \mathbb R^2 \to \mathbb R $$ that's smooth, we can ask "at the point $P$, what's the directional derivative of $f$ in the direction $v$?" This gives us a function -- let's call it $Q$ just to have a name, define by $$ Q(f) = \text{directional derivative of $f$ at $P$ in direction $v$} $$ which is defined on the set of differentiable functions. The function $Q$ has some nice properties: $Q(cf) = c Q(f)$ (where $c$ is a constant), and $Q(f + g) = Q(f) + Q(g)$, so $Q$ is actually linear. Both of these can be proved with fairly basic calculus. Slightly more surprising is that $$ Q(fg) = f(P) Q(g) + g(P) Q(f) $$ which is proved with the product rule for derivatives.

Whenever we have a function with these properties, we call it a 'derivation.'

Here's the first small surprise: not only can we take any vector at $P$ (not just $v$) and define such a derivation, but every derivation at $P$ actually turns out to be a derivation that comes from some vector.[This actually takes a bit of proof!]

So "tangent vectors at $P$" and "derivations at $P$" are in 1-1 correspondence.

What's nice about that is that when we talk about more general shapes, there might not be an obvious way to write down a tangent vector, but we can still think about derivations -- functions that have certain properties. And then we can say that the tangent space at $P$ is just the set of all "derivations at P".

Now what about curves? Every differentiable curve $\gamma$ with $\gamma(0) = P$ has a derivative $\gamma'(0)$ which is a "vector at $P$". But multiple curves can have the SAME derivative at $0$. For instance $$ \alpha(t) = (t, 0) $$ and $$ \beta(t) = (t, t^2) $$ both have $(1, 0)$ as their derivative at $t = 0$. So we make a rule that we're going to treat any two such curves as "equivalent". That divides the set of curves through the origin (or those through $P$) into a bunch of piles: the ones whose derivative is $(1, 0)$; the ones whose derivative is $(-3, \pi)$; the ones whose derivative is $(-11, -11)$, and so on.

We call these piles "equivalence classes", and it now becomes clear that such equivalence classes are in 1-1 correspondence with vectors. So we can forget vectors and talk about equivalence classes of curves instead. Again, not terribly helpful in the plane, but really nice when you want to talk about some infinite-dimensional manifold that isn't yet 'embedded' in a nice space like Euclidean space.

Summary: there are many different sets whose elements are in 1-1correspondence with tangent vectors at $P$. Which one you decide to call "tangent vectors" depends on your approach, and after a differential geometry/topology course or two, you find yourself switching among them pretty liberally.

John Hughes
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  • Wow, thank you so much, this is very clear, especially the summary. So the "tangent line" in $\mathbb{R}^2$ is just one way to represent the tangent vector, right? And if so, can we use the image of a chart to represent a tangent space? – wut Dec 02 '24 at 05:15
  • And I would also want to ask, how do we construct such "derivation" for a certain manifold because such a function seems very general? Is there any condition to guarantee a construction is valid? – wut Dec 02 '24 at 05:37
  • "The tangent line in R^2...": almost. I'd say "the tangent ray", because I want to keep the sense of direction, or even "the tangent arrow", so that I have direction and magnitude. And yes, we can actually define all this stuff in the image of a chart, and use the tangent vectors in that image (nice! because we're in $R^n$!) as tangent vectors to the manifold. The problem is where charts overlap: are the vectors in the image of chart1 the tangent vectors, or the ones in the image of chart2? The answer is to build an equivalence relation on these things. In classical differential geometry... – John Hughes Dec 02 '24 at 14:48
  • that gets expressed by saying "coordinates of tangent vectors transform according to the law XYZ", where XYZ is some messy thing involving sums and lots of indices. It took me years to figure out that was what was going on. :( – John Hughes Dec 02 '24 at 14:49
  • How to construct derivations? That's also typically done chart-wise. But I can't explain it in a 600-character comment. :( – John Hughes Dec 02 '24 at 15:13
  • Thank you so much! I am working on some manifolds that are not n-fold. Can I define another chart, but not to R^n but to like L_2(P)? And how do I check if a construction is valid? – wut Dec 02 '24 at 15:59
  • Uh...I don't even understand the question. A manifold, by definition, is locally homeomorphic to Euclidean space of some dimension. Anyhow, I feel as if I'm out of my depth at this point. – John Hughes Dec 02 '24 at 17:26
  • Oh, I see. Thank you very much anyway. You have helped me a lot! – wut Dec 02 '24 at 18:25