Recent content by Antarres

Yeah, I'm aware of that, quadratic equation was just the simplest example I could think of. For example, if ##f(x) = x^2 - ax + b##, and ##f(x)=0## is a surface defined through this equation, then for particular ##a## and ##b## this function will have a double root. In that case ##\sqrt{f(x)}...

@martinbn In case of Kerr solution, the horizon is defined in the coordinates in the link by ##\Delta=0##, and this is a quadratic function in the radial coordinate. In the extremal case, it has a double root. Taking the gradent(or just exterior derivative, to make it simpler), you'll find it's...

I'm not sure about your example of a bump function, that is, I don't understand what you mean by piecing it together to put up the function ##g##. I understand that principally, you don't have to define the normal vector with respect to a function, but this function defines a surface, and it's...

I'm not sure this is a differential geometry question, but I think it is. In general when we have a hypersurface(or in case of 3D space just a surface) it is defined with an equation ##f(x^a)=0## for some function ##f##. Then the normal vector is the gradient of this function, if we want an...

Well, to answer your last question, when you are making a projective transformation, what you're doing is that you're basically taking points in one space and using a congruence of lines(usually straight lines, that you can imagine as rays) you're connecting points from one space to the points...

@gnnmartin I'm not sure that I understand your construction, but when it comes to spherical coordinates on the earth(considered as a sphere), you don't need a third dimension to describe the terrain. You can use the same spherical coordinates to describe points on hills and valleys, since a...

@jbergman Well... I never said that you can make a single chart on a circle, I was just describing the way you extend your chart using Riemann normal coordinates around a point. Obviously you wouldn't be able to cover a whole manifold by a single chart.

The way to construct coordinates around a point is to look for ##n##(the dimension of the manifold), commuting vector fields which will be your coordinate basis. There's a statement in differential geometry that if you can find ##n## vector fields which all commute, then that is sufficient and...

All I'm saying is, ##\delta(0)## is independent of ##p## obviously, and your ##A(p)## is proportional to this, so it is also independent of ##p##. The question of being mathematically rigorous is basically about rewriting this in such a way so as to evade writing ##\delta(0)##, an undefined...

I'd say you're missreading your anticommutation relation in a sense. If you want to do this the physicist way(which means by being less formal about math) you would say: $$\{a^{r\dagger}_p, a^s_p\} = \delta^{rs}\delta(0)$$ This, as it stands clearly doesn't depend on ##p##, and so you have your...

What @Orodruin said is the correct explanation, but I will expand on that, in case you're still confused about this. It is true that on an ##n##-dimensional manifold the wedge product of 2-form is commutative. However, consider the specific case where a 2-form is a product of 2 basis...

I think I have a slight misconception maybe, but I was wondering about this question. Usually when we say that the vectors are parallel, we say that it means that there's an equation ##k = \alpha l##, for the vectors ##k## and ##l## and some scalar ##\alpha##. In the context of differential...

@samalkhaiat, this makes it a lot clearer to me. I will look at the links you provided for the details, but already the things you mentioned clarified a lot of what I was asking. Thank you!

@vanhees71 This is precisely what I was intending to say when giving the decomposition of the 2nd rank tensor under orthogonal group. You made it perfectly clear though, so thank you for that. I wanted, analogously to that, to look to explain the decomposition of the Riemann tensor, that is the...

Isn't a metric supposed to be symmetric? I guess there could be some generalization to make the metric have both symmetric and anti-symmetric parts, but the meaning of the metric in any case should be to measure distance(or I guess in case of metric tensor which acts on vectors, is to measure...