Hmmm. I don't see any obvious reason why that's true, so it's probably just a matter of working out the algebra. I'm going to guess, by analogy with the Euclidean case, that we have for some scalar function f. Then , so . Perhaps something magic happens and everything cancels?(Original post by latentcorpse)
oops. sorry. we want to show given that is orthogonal to the surface .
But actually, I think you've omitted a condition. I suspect you'll have to assume that or something like that in order to do this.

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 05012011 13:21
Last edited by Zhen Lin; 05012011 at 13:25. 
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 05012011 15:04
(Original post by Zhen Lin)
Hmmm. I don't see any obvious reason why that's true, so it's probably just a matter of working out the algebra. I'm going to guess, by analogy with the Euclidean case, that we have for some scalar function f. Then , so . Perhaps something magic happens and everything cancels?
But actually, I think you've omitted a condition. I suspect you'll have to assume that or something like that in order to do this.
http://www.maths.cam.ac.uk/postgrad/...07/Paper61.pdf
I don't see any extra conditions. though?
Why did you take ?
I've been reading through the notes and there's something about half way down p105 that I think might have something to do with this question but I'm not sure.
And one other thing about integration on manifolds. Why is there a in (344) when on the earlier pages where he defines integration on a manifold, he simply uses an action . I guess above (344), he does say define so should I just take it as it is without worrying about where it came from or do you happen to know?Last edited by latentcorpse; 05012011 at 15:31. 
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 05012011 16:41
(Original post by latentcorpse)
Well, here is the original question:
http://www.maths.cam.ac.uk/postgrad/...07/Paper61.pdf
I don't see any extra conditions. though?
Why did you take ?And one other thing about integration on manifolds. Why is there a in (344) when on the earlier pages where he defines integration on a manifold, he simply uses an action . I guess above (344), he does say define so should I just take it as it is without worrying about where it came from or do you happen to know?
gives you the scale factor needed to get a hypervolume that corresponds to intuition, i.e. if you have a small hypercube with side lengths , then as , its hypervolume . If you want to see an example to convince you why we need it — recall that for spherical polar coordinates for Euclidean 3space, the volume form is , and indeed . If you're still not convinced, remember that we get by computing the Jacobian determinant. It turns out that if you start from Cartesian coordinates, the determinant of the metric is exactly the square of the Jacobian determinant. (In fact, the metric is the "square" of the Jacobian matrix, in a sense.)
But yeah, it's basically by definition, but it's not completely unmotivated. 
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 05012011 18:23
(Original post by Zhen Lin)
In the Euclidean case, gives you the normal covector to the surface since a vector is tangent to the hypersurface iff . This should be familiar from multivariable calculus. Taking a literal reading of the question, I could take any multiple of that and still get a normal covector. 
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 05012011 18:38

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 05012011 21:08
(Original post by Zhen Lin)
Think about the definition of a level surface and what it means for a vector to be tangent to a level surface. This should be covered in any good multivariable calculus textbook.
But why is ? 
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 06012011 03:33
(Original post by latentcorpse)
Yeah so it's basically like saying that their inner product is 0 so they must be orthogonal and if is normal to the surface then must be tangent to the surface.
But why is ? 
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 06012011 11:41
(Original post by Zhen Lin)
Did I say it was? In all likelihood it's probably . But try it with for arbitrary f first and see where you get.
And if you could just confirm that for a vector field , and a scalar field , the Lie derivative is as follows
since by defn
by expanding in a coordinate basis
since covariant and partial derivatives are the same on scalar fields by definition.
This is now a scalar equation and so must be true in any basis so we can write it in abstract indices as
(this was the result I was after, I was just hopign you could confirm that my method is correct?)
Thanks!Last edited by latentcorpse; 06012011 at 15:12. 
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 06012011 16:07
And if you could just confirm that for a vector field , and a scalar field , the Lie derivative is as follows
since by defn
by expanding in a coordinate basis
since covariant and partial derivatives are the same on scalar fields by definition.
This is now a scalar equation and so must be true in any basis so we can write it in abstract indices as
(this was the result I was after, I was just hopign you could confirm that my method is correct?) 
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 06012011 18:26
(Original post by Zhen Lin)
If you accept that is a necessary and sufficient condition for to be tangent to the level surface, then this follows automatically. If you don't see why that's true, then go think about what the expression means. (It's a directional derivative. When is a directional derivative zero?)
That works. But I would just skip from to and then write . These are all basically different ways of writing the same thing, by definition.
Now I know that a cylinder is an intrinsically flat object but I want to prove this, i.e. find .
So I assume I use (382)? Is there a faster way than to work out all the Christoffel symbols and then compute teh Riemann tensor and substitute back into (382). Surely there must be?
I mean, he just writes on the 3rd line of the example "the metric is flat" like it's totally obvious! 
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 07012011 03:26
(Original post by latentcorpse)
I'm having a bit of trouble in the example on p114.
Now I know that a cylinder is an intrinsically flat object but I want to prove this, i.e. find .
So I assume I use (382)? Is there a faster way than to work out all the Christoffel symbols and then compute teh Riemann tensor and substitute back into (382). Surely there must be?
I mean, he just writes on the 3rd line of the example "the metric is flat" like it's totally obvious!
In other words, the fact that you can pick up a sheet of paper and roll it into a cylinder without introducing crinkles can be turned into a rigorous proof that the cylinder is flat. (Similarly, the cone is flat.) 
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 07012011 12:37
(Original post by Zhen Lin)
Sure, you can do that if you want. It's tedious. The usual way of doing it is to observe that there is an isometric (i.e. metricpreserving) diffeomorphism from (an open subset of) flat space to (an open subset of) the cylinder. Isometries preserve the various geometric invariants (e.g. curvature) so you conclude K = 0 on (that open subset of) the cylinder.
In other words, the fact that you can pick up a sheet of paper and roll it into a cylinder without introducing crinkles can be turned into a rigorous proof that the cylinder is flat. (Similarly, the cone is flat.)
On p115, in (394) he writes that
Why is that?
And at the end of (394) he has
Surely this is a typo and it should be ?
And in the remark at the top of p116, he says an n dimensional amnifold is maximally symmetric if it has n(n+1)/2 linearly independent Killing vector fields and he calims that this 5 dimensional de Sitter spacetime is maximally symmetric even though it only has 10 Killing vector fields (but form the formula it should have 15)?Last edited by latentcorpse; 07012011 at 12:58. 
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 07012011 18:25
Probably.And in the remark at the top of p116, he says an n dimensional amnifold is maximally symmetric if it has n(n+1)/2 linearly independent Killing vector fields and he calims that this 5 dimensional de Sitter spacetime is maximally symmetric even though it only has 10 Killing vector fields (but form the formula it should have 15)? 
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 07012011 21:10
(Original post by Zhen Lin)
I presume this de Sitter spacetime is the submanifold M he talks about in the preceding section. M is 4dimensional. Don't be confused by the fact that it's embedded in a 5dimensional manifold. (Analogy: the 2sphere can be embedded in as the boundary of the 3dimensional unit ball. But is nonetheless a 2dimensional surface.)
However, he says (again in that remark) that "just like 4d Minkowski spacetime, it is maximally symmetric". This suggests, 4d Minkowski spacetime is maximally symmetric. From the formula, there will be 4*5/2=10 linearly independent Killing vector fields. However, surely this time we'd be using a 4x4 antisymmetric matrix and so there would only be 6 independent parameters. Clearly 10 is not equal to 6! So I'm confused as to what the other four are: could it be that we get four from the fact that the Minkowski metric doesn't depend on any of t,x,y or z and then the other 6 come from the Lorentz transformations?
And just above (397), he says that the de Sitter metric (see eqn (396)) is a metric of constant curvature with . Do you know how to show that this is the value of K? I was going to use (382) but was wondering if there was a better way than having to calculate the Riemann tensor?Last edited by latentcorpse; 07012011 at 22:18. 
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 08012011 02:37
(Original post by latentcorpse)
So I guess this is kind of like saying that if we specify 4 of then the fifth is given by eqn (393) and hence there are 4 independent parameters for this de Sitter spacetime  meaning it is four dimensional. Correct?However, he says (again in that remark) that "just like 4d Minkowski spacetime, it is maximally symmetric". This suggests, 4d Minkowski spacetime is maximally symmetric. From the formula, there will be 4*5/2=10 linearly independent Killing vector fields. However, surely this time we'd be using a 4x4 antisymmetric matrix and so there would only be 6 independent parameters. Clearly 10 is not equal to 6! So I'm confused as to what the other four are: could it be that we get four from the fact that the Minkowski metric doesn't depend on any of t,x,y or z and then the other 6 come from the Lorentz transformations?
It's wellknown that a hyperboloid is a surface of constant curvature, like a sphere, but I don't know how that can be proven. 
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 08012011 12:44
(Original post by Zhen Lin)
Yes. Of course, this is a claim with some technical content in it and should be proven. (Recall the definition of dimension of a manifold requires you to check facts about the tangent spaces.)(Original post by Zhen Lin)
Yes — you forgot about translation symmetry. 
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 08012011 12:47
(Original post by latentcorpse)
I don't think this definition was in our notes. What would I have to check?So the Killing vectors correspond to four translational symmetries and 6 Lorentz symmetries then? 
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 01022011 21:48
(Original post by Zhen Lin)
Nevermind then. (The point is that the dimension of the tangent space is equal to the dimension of the manifold.)
I believe so.
The orbits ( integral curves ) of a vector field are the curves to which is everywhere tangent
where
Then it says this statement is equivalent to
Ok well the first statement kind of makes sense  it's saying that is in the same direction as the tangent vector, right?
I don't see how this is equivalent to the 2nd one though because if the second one were true and we were to use the fact that then
I'm 99% certain that what I just wrote in that last sentence is wrong because it would never be , it would always be but then I don't understand and was hoping you could sort out how we go from teh 2nd statement to the 1st or vice versa? Thanks! 
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 02022011 00:16
(Original post by latentcorpse)
Hey. Sorry to "reopen" this thread but I was wondering if you could clarify something in my notes for another class which also use differential geometry:
The orbits ( integral curves ) of a vector field are the curves to which is everywhere tangent
where 
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 06042011 12:44
(Original post by Zhen Lin)
I'm sorry, but this notation is incomprehensible. I can't even begin to guess what is meant here.
Thanks.
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