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I have the following Submanifold: $$M = \{(x,e^{x})\in \mathbb{R^{2}:x \in \mathbb{R}}\}$$ to which I have to find a tangent and normal space $T_{p}M$ and $N_{p}M$ at the point $p \in M$.

This is one of the examples that I have but in the same excercise I have like 10 different submanifolds to which i'll have to find $T_{p}M$ and $N_{p}M$. I am clear with their idea visually and we also had a Theorem about the Basis of $T_{p}M$ and $N_{p}M$.

if somebody could help me out with the practical solutions in these cases (since the excercises has like 10 different cases with different functions i assume that there might be a practical way to see them) i would be very thankful

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  • $\begingroup$ What did you try? $\endgroup$ Nov 22, 2020 at 21:26
  • $\begingroup$ i tried mostly to understand how i might do it. i saw the definitions and $T_{p}M$ is defined as the set of the Tangent vectors to $M$ in $p$ and then $NpM$ would be the transpose of $T_{p}M$. So about the tangent vectors I thought that maybe the first derivative would be 0 but i have no further clue how to present it formally or if this is right $\endgroup$
    – Annalisa
    Nov 22, 2020 at 21:38
  • $\begingroup$ $M$ is just the graph of a function, so you can find the tangent space just by finding the tangent line with calculus; then find the normal space by the relation between noraml and tangents. $\endgroup$
    – user147556
    Nov 22, 2020 at 21:38
  • $\begingroup$ @MichaelBarz so the normal space would be { $a$ ($\nabla f$, -1)} and then tangent space would be the transpose? $\endgroup$
    – Annalisa
    Nov 22, 2020 at 21:45

2 Answers 2

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Let $$M=\{(x,e^x): x\in \mathbb{R}\}$$ be a submanifold of $\mathbb{R^2}$. Let $\phi: \mathbb{R^2}\to \mathbb{R}$ defined by $\phi(x,y)= e^x-y$ on some open subset $U$ of $\mathbb{R^2}$. Notice $M$ is the level set $\phi^{-1}(0)$. Then since $T_p(M) = (\nabla f\vert p)^{\perp}$, then we can find the tangent space this way: $(\nabla f)=(e^x, -1)$. Therefore the complement is generated by $(1,e^x)$.

Notice that all tangent lines to $M$ in $\mathbb{R}^2$ are precisely $$span\{(1,e^x):x\in \mathbb{R}\}$$

If you have 50 similar exercises, I would follow this procedure.

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  • $\begingroup$ very helpful answer but i have two more questions. Why do you say that gradient of f is (e^x, -1) wouldn't it be (1, e^x)? And also what about the cases that i have many variables, do i only need to find the gradient? Or will there always be a 1 or -1? $\endgroup$
    – Annalisa
    Nov 23, 2020 at 10:08
  • $\begingroup$ @Annalisa The gradient is given by $(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y})$, which in this case is $(e^x,1)$. Same thing for several variables, just carefully write own $(\frac{\partial \phi}{\partial x_1}, \frac{\partial \phi}{\partial x_2}, \frac{\partial \phi}{\partial x_3},...,\frac{\partial \phi}{\partial x_n})$ $\endgroup$ Nov 23, 2020 at 17:14
  • $\begingroup$ how and why did you define the function e^x-y like this, in this case? $\endgroup$
    – Annalisa
    Jan 21, 2021 at 17:56
  • $\begingroup$ @Annalisa Sorry for the late reply, I was away from here for a bit. I defined a function of 2 variables $\varphi(x,y)=e^x-y$ in order to "capture" the exponential function $y=e^x$ as a level curve at $z=0$ $\endgroup$ Feb 3, 2021 at 3:15
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You can draw a picture and see what we expect:

enter image description here

Here is a rigorous calculation, using curves and derivations: Let $p=(x_0,e^{x_0})\in M$. Define $\gamma:[-\epsilon, \epsilon]\to M$ by $\gamma(t)=(t+x_0,e^{t+x_0}).$ Then, $\gamma(0)=p$ and $\gamma'(0)=a\frac{\partial}{\partial x}+b\frac{\partial }{\partial y}$ is an arbitrary tangent vector at $p$. Operating with $\gamma'(0)$ on the projections $r_1(x,y)=x$ and $r_2(x,y)=y,$ respectively, we get $\gamma'(0)r_1=a$ and $\gamma'(0)r_2=b$.

Now, by definition of $\gamma'(0)$, we have

$a=\gamma'(0)r_1=\frac{d (r_1\circ \gamma)}{dt}|_{t=0}=\frac{d (t+x_0)}{dt}|_{t=0}=1$ and $b=\gamma'(0)r_2=\frac{d (r_2\circ \gamma)}{dt}|_{t=0}=\frac{d (e^{t+x_0})}{dt}|_{t=0}=e^{x_0}$.

Therefore, in terms of components, $v\in T_pM$ is given by vectors in $\operatorname {span}\{(1,e^{x_0})\}.$

And then $N_pM$ is just the orthogonal complement, so we need to solve $A\cdot 1+B\cdot e^{x_0}=0.$ We know that this subspace has dimension $1$ (because $\mathbb R^2$ has dimension $2$) so all we need is one vector satisfying this condition. Insepction gives the vector $(e^{x_0},-1)$ and so we may take $N_pM=\operatorname {span}\{(e^{x_0},-1)\}.$

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