# Tangent Space to a manifold

So, I have a manifold $M=\{\mathbf{x}:\mathbf{\Theta}\left(\mathbf{x}\right)=\mathbf{0}\}$. I can also write $M=\{\mathbf{x}:\mathbf{F}(\mathbf{x})=\mathbf{c}\}$. Both functions are differentiable. I understand why we can see the Tangent space $T_{\mathbf{x}_0}$ as the $Ker(D\mathbf{\Theta}(\mathbf{x}_0))$.( I think...) But why can we not say the same about $Ker(D\mathbf{F}(\mathbf{x}_0))$? If we can, is $T_{\mathbf{x}_0}$ tangent to $\mathbf{\Theta}(\mathbf{x}_0)$ the same as $T_{\mathbf{x}_0}$ tangent to $\mathbf{F}(\mathbf{x}_0)$?Why?

To the first question, I would answer yes, and to the second no. Am I correct?

Many Thanks.

• This is too long for a comment, but is probably not enough to answer your question. In your first paragraph, it is a bit unclear (at least to me) what the issue is. It seems to me that you are suggesting that your manifold $M$ is cut out as the level set of a function from some larger manifold. If this is the case, then I don't understand the distinction between your two definitions of $M$ (i.e. what is the difference between $\mathbf{\Theta}$ and $\mathbf{F}$?). – THW Mar 30 '14 at 0:08
• Additionally, there seems to be some confusion regarding the tangent space in question. If you are trying to find the tangent space to a point $\mathbf{x}_{0} \in M$ and $M$ is cut out as a level set of some function defined on a larger ambient manifold, then $T_{\mathbf{x}_0}M$ is related to the kernel of $D\mathbf{F}(\mathbf{x}_0)$, not the image of $D\mathbf{F}(\mathbf{x}_0)$. Try and make the conditions on $M$, $\mathbf{F}$ and $\mathbf{\Theta}$ clearer so that we might be able to help address the question. – THW Mar 30 '14 at 0:08
• @THW You're right, I've just noticed my confusion on the third question. I've already removed it. – An old man in the sea. Mar 30 '14 at 0:13
• As to the first of your comments, the manifold can be described as the level set of two different functions, both differentiable. – An old man in the sea. Mar 30 '14 at 0:14
• The question still doesn't make sense though. Why do you think the tangent space can be viewed as the image of the derivative? – JLA Mar 30 '14 at 0:15

Assuming that you are viewing $M$ as the level set of a constant rank function $\mathbf{\Theta}$ defined on a larger ambient manifold . . .

For your question concerning why we (might) have $T_{\mathbf{x}_0}M = Ker\left(D\mathbf{\Theta}\left(\mathbf{x}_0\right)\right)$, you have to be a bit careful with the condition of $D\mathbf{\Theta}$. Namely, to ensure that $T_{\mathbf{x}_0}M = Ker\left(D\mathbf{\Theta}\left((\mathbf{x}_0\right)\right)$ you need to know that $D\mathbf{\Theta}$ is surjective at $\mathbf{x}_{0}$.

For a simple example of this, consider the following situation:

Let $M = \left\{ \mathbf{x}=(x, y) \in \mathbb{R}^{2} \Big\vert\, x = 0\right\}$. $M$ is clearly a submanifold of $\mathbb{R}^2$ and it is cut out by the function $\mathbf{\Theta}: \mathbb{R}^{2} \to \mathbb{R}$ defined by $\mathbf{\Theta}\left(\mathbf{x}\right) = x$. In this instance, $T_{\mathbf{x}_0}M = Ker\left(D\mathbf{\Theta}\left(\mathbf{x}_0\right)\right)$ for all $\mathbf{x}_{0}$ in $M$. However, you could also define $M$ as the level set of the function $\mathbf{F} : \mathbb{R}^{2} \to \mathbb{R}$ defined by $\mathbf{F}(\mathbf{x}) = x^2$. In this instance, $\mathbf{F}$ is not a map of constant rank as the rank drops at $\mathbf{x}_{0} = \left(0, 0\right)$ and $T_{\mathbf{x}_0}M \ne Ker\left(D\mathbf{F}\left(\mathbf{x}_0\right)\right)$, but instead $T_{\mathbf{x}_0}M \subset Ker\left(D\mathbf{F}\left(\mathbf{x}_0\right)\right)$.

Regarding the intuition of this situation, there is in fact a beautiful (and wonderfully intuitive) description. You have defined $M$ as a level set of some function $\mathbf{\Theta}$, which means that $\mathbf{\Theta}$ is constant along $M$. If $\mathbf{x}_0 \in M$ and $v \in T_{\mathbf{x}_{0}}M$ is tangent to $M$, then one expects $D\mathbf{\Theta}\left(\mathbf{x}_0\right)\left(v\right)$ to be zero precisely because $\mathbf{\Theta}$ is constant on $M$! (Note by the example above, $D\mathbf{\Theta} \left(\mathbf{x}_0\right)\left(v\right)$ being zero is merely a necessary condition for $v$ to be tangent to $M$ and not a sufficient condition for $v$ to be tangent to $M$.)

• Many thanks! By the way, do you know of a book, that gives the intuition, and that would allow me to self-study differential geometry? – An old man in the sea. Mar 30 '14 at 11:08
• So, if $D\mathbf{F}(\mathbf{x}_0)$ and $D\mathbf{\Theta}(\mathbf{x}_0)$ had the same rank, then $T_{\mathbf{x}_0}M=Ker(D\mathbf{F}(\mathbf{x}_0))$? – An old man in the sea. Mar 30 '14 at 11:30
• @ An old man in the sea, Re: The question of rank, you need the rank of $D\mathbf{F}$ (or $D\mathbf{\Theta}$) to be full (or maximal) at a point $\mathbf{x}_0 \in M = \mathbf{F}^{-1}\left(\mathbf{c}\right)$ in order to ensure that $T_{\mathbf{x}_0} M = Ker\left(D\mathbf{F}\left(\mathbf{x}_0\right)\right)$. – THW Mar 30 '14 at 19:16
• To be a bit more precise let $N$ be an $n$-dimensional manifold and $\mathbf{F}: N \to \mathbb{R}^{s}$ (w/ $n \ge s$) . If $\mathbf{c} \in \mathbb{R}^{s}$ is a regular value of $\mathbf{F}$, then $M = \mathbf{F}^{-1}\left(\mathbf{c}\right)$ is an $n - s$ dimensional submanifold of $N$ and for $\mathbf{x}_{0} \in M$ we have $T_{\mathbf{x}_{0}}M = Ker\left(D\mathbf{F}\left(\mathbf{x}_{0}\right)\right)$. In more generality, you can replace $\mathbb{R}^{s}$ in the above with an $s$-dimensional manifold. – THW Mar 30 '14 at 19:29
• Re: Books that allow for self-study and provide intuition, I would recommend starting with John Milnor's "Topology from a Differentialable Viewpoint", "Differential Topology" by Guilleman and Pollack, Michael Spivak's "Comprehensive Introduction to Differential Geometry," and/or John Lee's "Introduction to Smooth Manifolds." – THW Mar 30 '14 at 19:39