# Why are parallelizable manifolds called parallelizable?

Let $$M$$ be a smooth manifold of dimension $$n$$. Then $$M$$ is called parallelizable if there exists a global frame, i.e., $$n$$ linearly independent smooth vector fields $$X_1,X_2, \dots, X_n$$.

Why is a paralleizable manifold called "parallelizable"?

I guess it's related to the parallel transport on $$M$$. We can always choose a Riemannian metric $$g$$ on $$M$$ with the corresponding Levi-Civita connection $$\nabla$$ on the tangent bundle $$TM$$. Then we can define the parallel transport of a vector $$v \in T_pM$$ at any point $$p \in M$$ along any path $$C$$ starting from $$p$$.

Suppose $$M$$ is connected. Is it true that if $$M$$ is parallelizable, then we can always obtain a global frame via the parallel transport of a local frame $$v_1, \dots, v_n$$ of $$T_pM$$ at any point $$p \in M$$?

• If $M = S^3$,then $M$ is parallelizable. Let's give $M$ the round metric, just for definiteness. Then I worry about parallel translating a frame because that requires choosing a bunch of geodesics. I mean, given $p$ and any point $q\neq -p$, there is an essentially canonical choice of geodesic from $q$ to $p$ (the unique minimal one), but then what do you do if $q = -p$? (Of course, this is not a proof that it can't be done.) – Jason DeVito Oct 4 '19 at 2:19
• @JasonDeVito Why do we have to choose geodesics for parallel transport? – Yuhang Chen Oct 4 '19 at 3:26
• @YuhangChen Because parallel transport is defined using curves. They need not be geodesics, but the choice of route matters on any curved manifold. – Joonas Ilmavirta Oct 4 '19 at 3:46
• @YuhangChen: As Joonas said, they don't have to be geodesics, but then the choice of curve becomes even more arbitrary, which makes continuity of the frame even less likely. (Again, this is nothing close to a proof that what you're saying can't be done.) – Jason DeVito Oct 4 '19 at 13:26

It's not about parallel transport but the ability to reasonably define when two vectors on different tangent spaces are parallel. The vector fields provide a natural basis on each tangent space and allows us to compare two tangent vectors by comparing their components in this basis. With this you can decide when two vectors are equal or parallel.

If you are familiar with bundles, parallelizability means that the tangent bundle is globally trivializable. This is just another way to say that you can describe tangent vectors irrespective of the base point. In other words, for any two points $$x,y\in M$$ there is a canonical linear isomorphism $$F_{x,y}:T_xM\to T_yM$$ which maps each $$X_i(x)$$ to $$X_i(y)$$. (This property defines $$F_{x,y}$$ uniquely.)

A connection on the tangent bundle gives a way to identify tangent spaces along a curve, but not between any two points. Parallel transport is defined along curves, not between points. Parallelizability in the sense of this question has nothing to do with connections. The global trivialization of the tangent bundle gives you a way to naturally translate tangent vectors between base points, but it is not parallel transport in the usual sense.

The definition of parallelizability involves no metric or connection, so parallel transport doesn't even have enough structure to work in. You can always get a metric by defining your frame to be orthonormal. This gives a "global inner product", but the vectors are not generally parallel transported with respect to this metric.

The term "parallel" in the context of the question's definition must be understood in the sense of parallel vectors in a general vector space, not in the sense of parallel transport along curves.

• Shouldn't we require some connection on the tangent bundle in order to talk about whether two vectors are parallel or not? – Yuhang Chen Oct 4 '19 at 3:52
• @Yuhang: Within a single vector space, "parallel" makes sense: $v_1$ and $v_2$ are parallel if $v_1$ is a scalar multiple of $v_2$ or $v_2$ is a scalar multiple of $v_1$. Of course, if $v\in V$ and $w\in W$, it doesn't make sense to ask if they are parallel. But a choice of trivialization of the tangent bundle (if it exists) is essentially a consistent choice of isomorphism $f_{pq}:T_p M\cong T_q M$ all around the manifold. Once you have these isomorphisms in mind, asking that $v\in T_p M$ and $w\in T_q M$ be parallel reallys means asking that $f_{pq}(v)$ is parallel to $w$. – Jason DeVito Oct 4 '19 at 13:30
• @JasonDeVito What do you mean by a consistent choice of isomorphism $f_{pq}: T_pM \cong T_qM$ all around the manifold? It sounds like the "parallel translates" in John's answer. – Yuhang Chen Oct 4 '19 at 14:34
• @YuhangChen I updated the answer. To answer your first comment: No, no connection is needed. A connection will help identify tangent vectors along a curve, not generally between any two points. The definition of parallelizability in your question has nothing to do with parallel transport. The "parallel transport" John Hughes gives in his answer is typically not the parallel transport of any connection, but an entirely different kind of transport. – Joonas Ilmavirta Oct 4 '19 at 22:08
• The answer makes perfect sense. I agree that being "parallelizable" should be an intrinsic property of the smooth manifold $M$ considered, so it is not directly related to the notion of "parallel transport", which depends on the metric we choose on $M$. – Yuhang Chen Oct 4 '19 at 23:24

I think it's simpler than that -- you can define parallel transport of the vector $$v = a_1 X_1(p) + \ldots a_n X_n(p)$$ in $$T_p M$$ to be $$v' = a_1 X_1(q) + \ldots a_n X_n(q)$$ in $$T_q M$$. These "parallel translates" of $$v$$ would, if everything were just $$\Bbb R^n$$, with $$n$$ standard-unit-vector vector fields, be actually parallel vectors.

As for your last question: I suspect the answer is "no" -- all you have to do is to pick a metric that's a little wonky. But I don't have the heart to write an actual example right now.

• This is interesting. So we have an alternative (non-equivalent?) definition of "parallel transport" when the manifold $M$ is parallelizable. The last question seems trickier than I thought. A particular example I have in mind is $S^1$, on which I can make sense of "parallelizable" via the parallel transport of any tangent vector at any point of $S^1$. – Yuhang Chen Oct 4 '19 at 3:38
• It seems your definition of "parallel transport" gives a positive answer to my last question. – Yuhang Chen Oct 4 '19 at 4:01
• What I meant was that if you have a notion of parallel transport, I think ]that you can (under some not too strong conditions) derive a metric for which this is the parallel transport (but I'm by no means sure of that!). But you can also have a metric for which it is not, in which case the answer to your last question is likely to be "no." I suspect it's not hard to build an example on the torus, for instance, or even the plane! – John Hughes Oct 4 '19 at 10:39
• @John: I think it's not too hard to build an example on any flat Riemannian manifold, because then parallel transport only depends on the homotopy class of the curve (relative to both end points). In particular, in $\mathbb{R}^2$ with the usual metric, choosing any (reasonably smooth) curves at all and parallel translating just gives the usual notion of parallel. If your Riemannian manifold is not flat, then parallel transport is sensitive to the curve itself. That's why I suggested $S^3$ initially in a comment. (But, I don't really know how to prove it's impossible, hence just a comment) – Jason DeVito Oct 4 '19 at 13:33