June 10

What is the dimension of the space of all lines in ${\displaystyle \mathbb {C} ^{n}}$? And how do we see it is really a complex manifold? When we only look at lines that pass through the origin, we get ${\displaystyle \mathbb {P} ^{n-1}}$ which is a complex manifold of dimension ${\displaystyle n-1}$, so my intuition is that since we vary the line over all n direction of ${\displaystyle \mathbb {C} ^{n}}$ we get dimension ${\displaystyle (n-1)+n=2n-1}$, but I am told the actual answer is ${\displaystyle 2n-2}$. --אדי פ' (talk) 14:38, 10 June 2016 (UTC)[reply]

It depends on what you mean by a line. If you mean a 1-dimensional subspace of Cⁿ then the so called line is really a copy of the complex plane. The space of these is a real manifold of dimension 2n-2 but it's better to think of it as a complex manifold (complex projective space) of dimension n-1. If you mean line as a copy of the real line then you're basically throwing away the structure of Cⁿ as a complex vector space and viewing it as a real vector space of dimension 2n. The space of lines in this case is a real manifold of dimension 2n-1. The terminology is confusing but simpler of you understand that complex dimension is twice the real dimension. For example the complex projective line is really the Riemann sphere, and the complex projective plane is actually a manifold with 4 real dimensions. --RDBury (talk) 15:35, 10 June 2016 (UTC)[reply]

The projective space is the space of lines through the origin, not the whole space of lines, which are the affine one-dimensional subspaces. One way to get the dimension is to take a pair of hyperplanes A and B in ${\displaystyle \mathbb {C} ^{n}}$. Generically, a line intersects each of A and B in a unique point, so ${\displaystyle A\times B}$ gives a system of local coordinates. Because each of A and B is a hyperplane, this has dimension ${\displaystyle 2(n-1)}$. This leaves out the lines that are parallel to A or B. Going from one system of coordinates to another, defined by a different pair ${\displaystyle A',B'}$ of hyperplanes, the transition functions of the local coordinate system are holomorphic (in fact, fractional linear), so they patch together to a complex manifold. It is in fact the case that the space of lines is a complex algebraic variety. Indeed, if we realize ${\displaystyle \mathbb {C} ^{n}}$ as the set of finite points in ${\displaystyle \mathbb {CP} ^{n}}$, then the space of lines is the open subset of the Grassmannian ${\displaystyle Gr(2,\mathbb {C} ^{n+1})}$ of two-dimensional linear subspaces of ${\displaystyle \mathbb {C} ^{n+1}}$ that do not contain a line at infinity. This is an open condition, and the Grassmannian is known to be an algebraic variety via the Plücker embedding.

Biały 15:53, 10 June 2016 (UTC)[reply

]

When I said space of lines I meant the space of lines through the origin. This is the standard way of defining projective space and what the original question was referring to. --RDBury (talk) 16:16, 10 June 2016 (UTC)[reply]

No, the original question was the space of all lines, not just those through the origin. This is a complex manifold of complex dimension 2 (n-1). It is also a real manifold of dimension 4 (n-1), but that was not the question.

Biały 15:35, 11 June 2016 (UTC)[reply

]

Yes, on reading the question another few times I see I was misreading it. --RDBury (talk) 04:51, 12 June 2016 (UTC)[reply]

I think that it may be helpful to point out more specifically where the initial intuition is led astray. Sławomir does explain the derivation afresh, which does not clearly highlight this. To "vary the line over all n direction of ${\displaystyle \mathbb {C} ^{n}}$" corresponds to the translation of the origin through space. Translations parallel to the line does not produce new lines, and hence the reduction by 1 in the dimensions added by translations. A similar argument would reduce the dimension of the space of planes due to translations by 2 instead. —Quondum 19:51, 12 June 2016 (UTC)[reply]

Thank you all, it was very helpful.--אדי פ' (talk) 01:05, 13 June 2016 (UTC)[reply]

The Radian and Steradian are dimensionless units for the "surface area" (dimension d-1) of the Circle and the Sphere. Are there similar units for higher dimensional spheres?Naraht (talk) 18:00, 10 June 2016 (UTC)[reply]

I know of no specific name, but the definition ${\displaystyle A/r^{n-1}}$ for n-dimensional space, where A denotes the n-dimensional surface area of the subset of the n-dimensional sphere in question, seems to generalize very naturally.--Jasper Deng (talk) 19:34, 10 June 2016 (UTC)[reply]

You can use rad^d−1 (sr = rad²). I don't think the higher powers have special names. ("Hypersteradian" does get a few Google hits, but not many.) -- BenRG (talk) 19:58, 10 June 2016 (UTC)[reply]

I understand that if a function satisfies certain conditions, specifically it is cyclic, and maybe some others, then it can be expressed as the sum of a series of sine functions. I can learn much more about this by reading Fourier analysis.

I vaguely recall a similar concept. If a function satisfies some other set of conditions (essentially, it's zero-valued or otherwise boring except in one neighbourhood), then it can be expressed as the sum of a series of exponentials. Did I just dream this? Or if it's so, where can I read about it? Maproom (talk) 19:05, 10 June 2016 (UTC)[reply]

Well technically any function that can be expressed as the sum of a series of sine functions can be expressed as a series of exponentials using the identity ${\displaystyle \sin x={\frac {e^{ix}-e^{-ix}}{2i}}}$ 2001:630:12:2428:A016:2B2E:370F:6610 (talk) 19:23, 10 June 2016 (UTC)[reply]

If you look at the picture sideways you could say that if a function is periodic (and with other nice properties like continuity) with period 2πi then it's the sum of a series of functions of the form e^nx, where n ranges over the integers. There's also the Fourier transform which uses an integral instead of a sum and may be closer to the idea in the original question. --RDBury (talk) 19:46, 10 June 2016 (UTC)[reply]

The "otherwise boring except in one neighborhood" condition may be that it is a "Schwarz function" in the Schwartz space. The Fourier transform is an automorphism of this space, which is used to define Tempered distributions & define the Fourier transform for them.John Z (talk) 23:44, 10 June 2016 (UTC)[reply]

I think you want Laplace transform which, rather loosely speaking, is to the exponential function what the Fourier transform is to the sine wave. — Preceding unsigned comment added by 82.46.116.9 (talk) 19:03, 12 June 2016 (UTC)[reply]

My thanks to all of you. I now have enough leads to find out all I want to know. Maproom (talk) 08:05, 13 June 2016 (UTC)[reply]

compact support is the term for a function that is only non-zero in a sort of finite neighborhood. SemanticMantis (talk) 15:53, 13 June 2016 (UTC)[reply

]

Strictly speaking, compactness is stronger than boundedness.--Jasper Deng (talk) 20:45, 13 June 2016 (UTC)[reply]