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Topic: Lovelock and Rund: Star shaped set of points on a manifold
Replies: 7   Last Post: Jan 17, 2013 2:39 PM

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quasi

Posts: 10,226
Registered: 7/15/05
Re: Lovelock and Rund: Star shaped set of points on a manifold
Posted: Jan 16, 2013 3:46 PM
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quasi wrote:
>Hetware wrote:
>>This is from Lovelock and Rund, _Tensors, Differential Forms,
>>and Variational Principles_, pg. 142.
>>
>>"[f]or a given point P on X_n let us choose our coordinates
>>such that x^1=...=x^n=0 at P, after which we construct an
>>open set U on X_n which is defined by the property that for
>>any point Q element of U with coordinates x^h, the segment
>>consisting of the points with coordinates tx^h, 0<=t<=1, is
>>also contained in U."
>>
>>I'm having a bit of trouble grasping that concept. Let's
>>take R^2, for example. I choose a point in the middle of my
>>paper, and call it {0,0}. I now declare it to be a member of
>>some open set U.

>
>You mean, let U be an open set containing (0,0).
>

>>The smallest possible U is some infinitesimal open disk
>>centered on P.

>
>There's no smallest possible U.
>
>Perhaps you meant to say something like:
>
>U can be chosen as an open disk centered at (0,0) with an
>arbitrarily chosen positive radius, regardless of how small.
>

>>If I chose some point in U (whatever that means)
>
>In this context it means that the property to be discussed
>holds for all points of U.
>

>>and multiply it by some t from the closed interval [0, 1], I
>>get an infinitesimal curve segment including P and Q.

>
>Since you are in R^2, you get the line segment from P to Q,
>that is, the line segment from (0,0) to Q.
>
>There's nothing infinitesimal about it.
>
>The length of the segment depends on the choice of Q, but it's
>an actual line segment (degenerate if Q = (0,0).
>

>>So am I correct in understanding that the "star shaped" set U
>>is the infinitesimal disk centered on P, with parametrized
>>curves radiating infinitesimally from P?

>
>No.
>
>U is star-shaped in the sense that any point in U is "visible"
>from (0,0).
>
>Thus, U need not be a disk.
>
>For example, any convex open set containing (0,0) qualifies.
>
>But U need not be convex either.
>
>For example, the interior of a Jewish star centered at (0,0)
>also qualifies.
>
>Staying with the religious theme, the interior of a Christian
>cross centered at (0,0) qualifies as well.
>
>Perhaps now you get the picture.
>

>>If I follow a constructive approach, it seems that U cannot
>>have a finite extension.

>
>I have no idea what you mean by that.
>

>>If U were chosen such that some x^h > 1 at some point Q in U,
>>the whole thing would explode, since both my coordinate
>>magnitude and my parameter could be greater than 1, thus
>>identifying points with yet larger coordinate magnitudes.

>
>No, the parameter t is required to satisfy 0 <= t <= 1, so
>the points t*x^h stay on the line segment from (0,0) to Q.
>

>>Is any ball with x^i < 1 a permissible choice for U?
>
>An open ball of radius 1 centered at (0,0) is a legal choice
>for U, but the concept of
>
> "an open star-shaped region containing (0,0)"
>
>allows for lots of other possible shapes


It occurs to me that your confusion may be based on a
misinterpretation of the book's notation.

Many books on Differential Geometry use the notation

x^i

to refer to the i'th coordinate of the point x.

For example, in R^3, a point x would be represented as

(x^1, x^2, x^3)

whereas in other areas of math, the notation

(x_1, x_2, x_3)

would be used instead.

Thus, in the context of your discussion, x^h does _not_
mean x to the h'th power. It just means x_h, the h'th
coordinate of x.

If that was the cause of your confusion, it means you probably
failed to read the early pages of the book where I'm fairly
sure the authors specified that notational choice.

quasi



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