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In page-49 of MTW (1973 edtn), the following picture is shown:

enter image description here

After seeing this picture, the question which arose in my head is why exactly can we not define a vector as difference of points in curved space?

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    $\begingroup$ How do you define a "difference of points" in curved space? $\endgroup$
    – d_b
    Jun 18, 2021 at 5:09
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    $\begingroup$ @d_b In affine space the operation $A=B+\vec{v}$ is defined, the result of operation $A-B$ is then defined as a vector which needs to be added to point $B$ to get point $A$ $\endgroup$
    – Umaxo
    Jun 18, 2021 at 5:17
  • $\begingroup$ Take two nearby points on the surface, take difference, why is that amibgous? @d_b $\endgroup$ Jun 18, 2021 at 5:19
  • $\begingroup$ @Umaxo Ok, but that is just affine spaces. That's a long way off from the general case of (pseudo-) Riemannian manifolds. $\endgroup$
    – d_b
    Jun 18, 2021 at 5:59
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    $\begingroup$ @Umaxo I guess we are heading off-topic in a meta discussion, but briefly, if someone asks "Why is X impossible? Why can't we just do Y?" I believe it's pedagogically useful to say "Ok, try to do Y and see what happens" so they can learn where Y breaks down or doesn't actually lead to X. Your example is opposite, but similarly you might ask "Why can't we compute the area under a graph? Couldn't we just break the area up into pieces and count the pieces?" and a useful answer would be "You can! Now sit down and try to work out how to actually do that in practice." $\endgroup$
    – d_b
    Jun 18, 2021 at 6:22

3 Answers 3

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The short answer is that vectors must satisfy the eight axioms of a vector space in order to qualify as vectors.

Flat space is professionally known as affine space, which I have described in my answer here. In affine space, one can subtract two points to obtain a vector. By defining an origin and position vectors, one can also make points behave like vectors, therefore making it a vector space.

However, on a manifold, the situation is quite different. Position vectors can no longer be defined because there is no way to make points add like vectors. Even if we smoothly label the points using coordinates $x^1, \ldots ,x^n$, where $n$ is the dimension, there is simply no way to choose coordinates such that points obey the vector space axioms. This is why the "difference of two points" is no longer a meaningful definition of a vector. Conversely, if the points in our manifold do end up behaving like vectors, our manifold is an affine space.

However, the partial derivatives along all smooth curves passing through a point on the manifold do form a vector space, which is the tangent space.

I believe this answer on Mathematics.SE (which I personally upvoted) gives a really good summary of the situation.

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  • $\begingroup$ I am not sure what precisely you mean "points add like vectors". You cannot add points even in affine space. You can also define origin and position vectors in curved space, for example in submanifold of lie group connected to unity by exponential map, the origin is unit element ($e$) and then you can define position vectors for which $A=e+\vec{v}$ using exponential map. Difference with affine space is that in affine space you can choose whatever origin, while in curved lie group you can not and in general need to stick with $e$. $\endgroup$
    – Umaxo
    Jun 18, 2021 at 7:12
  • $\begingroup$ @Umaxo If you read carefully, I said that points can be added just like vectors in affine space if an origin is specified. On a manifold, there is no guarantee that the exponential map works globally, just like how you can choose Riemann normal coordinates only locally. I think this answer on Mathematics.SE sums it up quite well. $\endgroup$ Jun 18, 2021 at 7:20
  • $\begingroup$ You are not adding points, but vectors or vectors to points. You cannot add points in general affine space, no matter the origin. ...about the second point I said submanifold of lie group connected to unity by exponential map, where there is a guarantee. Yes, it is a special case, but it is also curved and there is good definition of origin and position vectors, contrary to your claim. Its just not an affine space, because it works only for one origin and not for all. $\endgroup$
    – Umaxo
    Jun 18, 2021 at 7:31
  • $\begingroup$ @Umaxo An origin in affine space sets up an isomorphism between points and vectors, which also makes it a vector space. As for your second point, the exponential map is the whole idea behind the Lie algebra of a Lie group. It tells us how to reach a nearby point from an existing point (certainly not necessarily at the origin). This is why we only need to know the infinitesimal generators in order to cover the entire group using infinitesimal steps. As a side note, the OP is using MTW which focuses more on Riemannian manifolds. $\endgroup$ Jun 18, 2021 at 8:16
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For one thing, vectors obey linear structure, i.e. we can write $\vec{v}$ as sum of two vectors $\vec{v_1}$ and $\vec{v_2}.$

So if we want to have operation $A=B+\vec{v},$ we should also demand that operations $$A=B+\vec{v}_1+\vec{v}_2=(B+\vec{v}_1)+\vec{v}_2=C+\vec{v}_2$$ and $$A=B+\vec{v}_2+\vec{v}_1=(B+\vec{v}_2)+\vec{v}_1=D+\vec{v}_1$$ produce the same point.

In curved space however, going first in the direction of $\vec{v}_1$ and then $\vec{v}_2$ does not produce the same point as going first in the direction of $\vec{v}_2$ and then $\vec{v}_1$. The difference between these two paths actually defines Riemann tensor, which characterizes curvature of space and is zero only if space is flat.

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Because in curved spacetime you can't take the difference of a vector at point P and point P' unlike in Minkowski spacetime.

Imagine abovementioned two points separated by $dx^{\mu}$ and let's call the vector at point P $a_{\mu}(P)$ and the other $a_{\mu}(P')$. You can infinitesimally transport the vector at P by $dx^{\mu}$. Thus $A_{\mu} = a_{\mu} + \delta a_{\mu}$ which is a vector at point P'. Now you can take the difference, hence it becomes

$$a_{\mu}(P') - A_{\mu}(P') = (a_{\mu, \nu} - \Gamma^{\lambda}_{\mu \nu} a_{\lambda})dx^{\nu}$$

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