# Dimension of the tangent space

This question is regarding the dimension of the tangent space $T_p\mathbb{R}^n$ as it is defined in the context of smooth manifolds. One the one hand, $T_p \mathbb{R}^n$ can just be interpreted as a copy of $\mathbb{R^n}$ based at a chosen point $p \in \mathbb{R}^n$. Now, since this is just $\mathbb{R}^n$ translated through $p$, clearly $T_p \mathbb{R}^n$ has the same dimension as $\mathbb{R}^n$. To see this, just take the canonical basis $\{e_1, \dots, e_n \}$ translate it through $p$ and note that this will yield a basis of $T_p \mathbb{R}^n$. Thus, $dim( T_p \mathbb{R}^n) = n$

On the other hand though, for a given $p$, it is common to define the tangent space of $\mathbb{R}^n$ at $p$ as $T_p\mathbb{R}^n = \{(p,x) | x \in \mathbb{R}^n \}$ Here, $T_p\mathbb{R}^n$ is effectively a Cartesian product of the one point vector space $\{p\}$ and $\mathbb{R}^n$. By this reasoning,

$$\displaystyle dim(T_p\mathbb{R}^n) = dim(\{p\} \times T_p \mathbb{R}^n) = dim(\{p\}) + dim(\mathbb{R}^n) = 1 + n$$

Obviously, this last equation can't be correct but I'm not really sure where my reasoning is off. So, specifically, my question is where is the error in my logic?

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The dimension of a point is zero. – Qiaochu Yuan Jun 6 '11 at 20:52
Are you sure that $\{p\}$ is a vector space (if $p\neq 0$) and that its dimension is $1$? :) – t.b. Jun 6 '11 at 20:53
D'oh. For some reason, I had it fixed in my brain that one point subsets had dimension 1 since any one point subset can serve as a basis for, say, $\mathbb{R}$. Thanks for pointing out my sillyness! – ItsNotObvious Jun 6 '11 at 21:00
$(p,x)\mapsto x$ yiels a linear isomorphism from $\{(p,x) | x \in \mathbb{R}^n \}$ to $\mathbb{R}^n$, so they have the same dimension. – wildildildlife Jun 6 '11 at 22:02

After having convinced ourselves that a point should by all means have dimension $0$, we should still think a little bit more. I'm not going to answer the dimension question, as this seems to be settled by now.

There are actually two notions of tangent spaces, but in order to see it, we should look at a better example than $\mathbb{R}^n$ itself, so let us look at the $n$-sphere $$S^{n} = \{(x_0,\ldots,x_n)\in \mathbb{R}^{n+1}\,:\,x_{0}^2 + \cdots + x_{n}^2 = 1\},$$ which is only slightly more complicated for this issue.

• The tangent space at a point $p \in S^{n}$ is $T_p S^{n} = \{x \in \mathbb{R}^{n+1}\,:\, p \cdot x = 0 \} = p^{\perp}$. For instance, this can be seen by computing that $f(p) = \sum x_{i}^2$ has derivative $df(p)h = 2 p \cdot h$ and by definition $T_{p}S^{n} = \ker{df(p)}$. Note that this is a vector space (in particular it contains $0$). While it's not the space actually tangent to $S^{n}$ at the point $p$, it is this space that is usually called the tangent space (because vector spaces are very convenient to work with).

• The affine tangent space is the space $p + T_{p} S^{n}$ and is actually the plane tangent to $S^{n}$ at the point $p$. This is not a vector space, but an affine subspace of $\mathbb{R}^{n+1}$.

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+1 - because the two can easily be confused – Juan S Jun 6 '11 at 23:42
@Qwirk: Ah, thanks! I sort of remembered that I had said such things before... – t.b. Jun 6 '11 at 23:45
If you talk about the dimension of a point, one might remark this point can be thought of as a affine space of dimension $0$, in order be able to think about dimension. – shuhalo Jun 7 '11 at 1:06
Thanks for the example and clarification – ItsNotObvious Jun 7 '11 at 11:41