# The Unapologetic Mathematician

## The Tangent Bundle of a Euclidean Space

Let’s look at the tangent bundle to a Euclidean space. That is, we let $V$ be a finite-dimensional real vector space with its standard differentiable structure and see what these constructions look like.

Well, if $V$ has dimension $n$, then we know $V$ is an $n$-manifold, and thus we know that $\mathcal{T}_vV$ is an $n$-dimensional vector space for every point $v\in V$. Now, of course all $n$-dimensional vector spaces are equivalent, but I say that $\mathcal{T}_vV$ can be made canonically equivalent to $V$ itself.

First of all, let’s set up an isomorphism between $V$ and $\mathcal{T}_0V$. Given a vector $v\in V$, we can set up the curve $c(t)=tv$. Then we just define a vector in $\mathcal{T}_0V$ by picking the tangent vector to $c$ at $c(0)=0\in V$. This correspondence is clearly linear. For instance, if $v$ and $w$ are two vectors in $V$, then $tv+tw = t(v+w)$. The linearity of the derivative shows that the tangent vector to $t(v+w)$ is the sum of the tangent vectors to $tv$ and $tw$.

In terms of components, pick a basis $\{e_i\}$ of $V$ and use it to get a coordinate map on all of $V$. We also get a basis of coordinate vectors for $\mathcal{T}_vV$ at each point $v\in V$, and in particular at $v=0$. What does this isomorphism look in terms of these coordinates?

Well, the $i$th component of the tangent vector at $c(0)=0$ is the derivative of the $i$th component of the curve written out in terms of coordinates. And this component is $tv^i$, where $v^i$ is the $i$th component of $v$ in the $\{e_i\}$ basis, so the derivative in question is just $v^i$. That is, if we use a particular basis of $V$ and the basis of coordinate vectors it induces on $\mathcal{T}_0V$, we get the exact same components for each vector $v\in V$ and its corresponding vector in $\mathcal{T}_0V$.

In fact, there’s nothing particularly special about $0\in V$ here. We can do pretty much the exact same thing at any other point $p\in V$; just replace the curve $c(v)=tv$ with the curve $c(v)=p+tv$. That is, we use the same line translated (slid around) parallel to itself by adding an offset of $p$ to every point. This is such a fundamental technique in Euclidean spaces that we have a name for this method of comparing vectors in $\mathcal{T}_pV$ with vectors in $\mathcal{T}_0V$ — and thus with vectors in any other $\mathcal{T}_qV$ — “parallel translation”.

Parallel translation works as simply as this for two fundamental reasons: we can cover the entire manifold $V$ by a single coordinate patch, and the vector space structure allows us to sensibly define an operation of “add $p$ to each point”. In more general manifolds, this parallel translation operation is not possible, or at least not so straightforwardly. But Euclidean spaces are, of course, special.

In fact, they’re so special that they’re basically all that most people study in multivariable calculus. And in that situation everything works so smoothly that people often conflate many vector spaces which are really only equivalent. It’s common to automatically compare tangent vectors rooted at different points, parallel translating them with impunity. It’s even common to cavalierly identify the tangent space $\mathcal{T}_vV$ with the space $V$ itself, identifying a point $p$ with its position vector — that which points from $0$ to $p$.

But to move forward in differential geometry it is absolutely essential to unlearn this identification. When working in a Euclidean space, it’s okay to use it to simplify matters, but in general it pays off to be careful whether we’re talking about a point in a manifold, a tangent vector at one point in a manifold, or a tangent vector at a different point in the manifold. They’re just not all the same thing.

April 11, 2011 - Posted by | Differential Topology, Topology