# Find a $T_pM$ and $N_pM$

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

• What did you try? Nov 22, 2020 at 21:26
• 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 Nov 22, 2020 at 21:38
• $M$ 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.
– user147556
Nov 22, 2020 at 21:38
• @MichaelBarz so the normal space would be { $a$ ($\nabla f$, -1)} and then tangent space would be the transpose? Nov 22, 2020 at 21:45

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.

• 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? Nov 23, 2020 at 10:08
• @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 $(\frac{\partial \phi}{\partial x_1}, \frac{\partial \phi}{\partial x_2}, \frac{\partial \phi}{\partial x_3},...,\frac{\partial \phi}{\partial x_n})$ Nov 23, 2020 at 17:14
• how and why did you define the function e^x-y like this, in this case? Jan 21, 2021 at 17:56
• @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 $z=0$ Feb 3, 2021 at 3:15

You can draw a picture and see what we expect:

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 $$N_pM=\operatorname {span}\{(e^{x_0},-1)\}.$$