vision - Does retinal detachment happen more frequently at night?

In case of normal, healthy person, no, I don't think so. In fact, retinal detachment could happen at anytime of the day. Main causes are diseases and illness, such as AIDS, diabetic retinopathy, cancer, or trauma such as post-cataract surgery and being hit/kicked hard on the eyes.

evolution - Why do plants have green leaves and not red?

There are several parts to my answer.

First, evolution has selected the current system(s) over countless generations through natural selection. Natural selection depends on differences (major or minor) in the efficiency of various solutions (fitness) in the light (ho ho!) of the current environment. Here's where the solar energy spectrum is important as well as local environmental variables such as light absorption by water etc. as pointed out by another responder. After all that, what you have is what you have and that turns out to be (in the case of typical green plants), chlorophylls A and B and the "light" and "dark" reactions.

Second, how does this lead to green plants that appear green? Absorption of light is something that occurs at the atomic and molecular level and usually involves the energy state of particular electrons. The electrons in certain molecules are capable of moving from one energy level to another without leaving the atom or molecule. When energy of a certain level strikes the molecule, that energy is absorbed and one or more electrons move to a higher energy level in the molecule (conservation of energy). Those electrons with higher energy usually return to the "ground state" by emitting or transferring that energy. One way the energy can be emitted is as light in a process called fluorescence. The second law of thermodynamics (which makes it impossible to have perpetual motion machines) leads to the emission of light of lower energy and longer wave length. (n.b. wavelength (lambda) is inversely proportional to energy; long wavelength red light has less energy per photon than does short wavelength violet (ROYGBIV as seen in your ordinary rain bow)).

Anyway, chlorophylls A and B are complex organic molecules (C, H, O, N with a splash of Mg++) with a ring structure. You will find that a lot of organic molecules that absorb light (and fluoresce as well) have a ring structure in which electrons "resonate" by moving around the ring with ease. It is the resonance of the electrons that determine the absorption spectrum of a given molecule (among other things). Consult wikipedia article on chlorophyll for the absorption spectrum of the two chlorphylls. You will note that they absorb best at short wavelengths (blue,indigo,violet) as well as at the long wavelengths (red,orange,yellow) but not in the green. Since they don't absorb the green wavelengths, this is what is left over and this is what your eye perceives as the color of the leaf.

Finally, what happens to the energy from the solar spectrum that has been temporarily absorbed by the electrons of chlorophyll? Since its not part of the original question, I'll keep this short (apologies to plant physiologists out there). In the "light dependent reaction", the energetic electrons get transferred through a number of intermediate molecules to eventually "split" water into Oxygen and Hydrogen and generate energy-rich molecules of ATP and NADPH. The ATP and NADPH then are used to power the "light independent reaction" which takes CO2 and combines it with other molecules to create glucose. Note that this is how you get glucose (at least eventually in some form, vegan or not) to eat and oxygen to breath.

Take a look at what happens when you artificially uncouple the chlorophylls from the transfer system that leads to glucose synthesis. http://en.wikipedia.org/wiki/Chlorophyll_fluorescence Notice the color of the fluorescence under UV light!

Alternatives? Look at photosynthetic bacteria.

mp.mathematical physics - The Quantum Operations On The Bipartite Systems

Given two distinct and noninteracting quantum mechanical
systems $mathfrak{S}_1$ and $mathfrak{S}_2$ with state spaces
$mathcal H_1$ and $mathcal H_2$, respectively, the state space
of the combined system $mathcal S_1+mathcal S_2$ is the
tensor product Hilbert space
$mathcal H=mathcal H_1otimesmathcal H_2$. Density operators
$Winmathcal D(mathcal H)$, and effects
$Finmathcal E(mathcal H)$. Similarly, there are corresponding
symbols $W_iinmathcal D(mathcal H_i),
F_iinmathcal E(mathcal H_i)$ for subsystems
$mathfrak{S}_i(i=1,2)$, respectively.

Given any quantum operation, $Phi:
mathcal D(mathcal H)rightarrow mathcal D(mathcal H)$, of
the composite system $mathcal S_1+mathcal S_2$.

Problem: (1) Do there exist whether or not two quantum
operation $phi_1$ and $phi_2$, of the subsystems $mathfrak{S}_1$
and $mathfrak{S}_2$, respectively, such that the following
diagram is commutative:

$$
begin{diagram}
node{mathcal D(mathcal H_1)} arrow[4]{e,t}{phi_1}node[4]{mathcal D(mathcal H_1)}\
node{}\
node{mathcal D(mathcal H_1otimesmathcal H_2)}
arrow[2]{n,l}{Tr_2} arrow[4]{e,t}{Phi} arrow[2]{s,l}{Tr_1}
node[4]{mathcal D(mathcal H_1otimesmathcal H_2)} arrow[2]{s,r}{Tr_1} arrow[2]{n,r}{Tr_2}
\
node{}\
node{mathcal D(mathcal H_2)} arrow[4]{e,b}{phi_2}
node[4]{mathcal D(mathcal H_2)}
end{diagram}
$$
i.e.
$$begin{eqnarray}
Tr_2(Phi(W))&=&frac{tr(Phi(W))}{tr(phi_1(Tr_2(W)))}phi_1(Tr_2(W)),\
Tr_1(Phi(W))&=&frac{tr(Phi(W))}{tr(phi_2(Tr_1(W)))}phi_2(Tr_1(W)),
end{eqnarray}
$$
where $phi_i: mathcal D(mathcal H_i)rightarrow
mathcal D(mathcal H_i)(i=1,2)$ and $Tr_i:
mathcal D(mathcal H)rightarrow
mathcal D(mathcal H_i)$ is a partial trace with respect to the subsystem $mathfrak{S}_i(i=1,2)$.

(2) If quantum operation $phi_1$ and $phi_2$ exist, give the
relationship among the quantum operations $Phi, phi_{1}$ and
$phi_2$.

gr.group theory - Presentation for the double cover of A_n

Yeah, Schur did this a long time ago. Let $tilde Sigma_n to Sigma_n$ be a double cover (there are two) -- lets denote them $tilde Sigma_n = Sigma_n^epsilon$ where $epsilon in {+1, -1}$.

Schur uses the notation $[a_1 a_2 cdots a_k]$ for a specific lift of the cycle $(a_1 a_2 cdots a_k) in Sigma_n$ to $Sigma_n^epsilon$ -- might as well call these $k$-cycles. Then his presentation goes like this:

$$[a_1 a_2 cdots a_k] = [a_1 a_2 cdots a_i][a_i a_{i+1} cdots a_k] forall 1 < i< k$$

and all $k$-cycles, $k>1$.

$$[a_1 a_2 cdots a_k]^{[b_1 b_2 cdots b_j]} = (-1)^{j-1}[phi(a_1) phi(a_2) cdots phi(a_k)]$$

where $phi$ is the cycle $(b_1 b_2 cdots b_j)$

$$[a_1 a_2 cdots a_k]^k = epsilon$$

for all $k$-cycles -- ie this is always $+1$ or $-1$ depending on which extension of $Sigma_n$ you're interested in. And:

$$[a_1 a_2 cdots a_k][b_1 b_2 cdots b_j] = (-1)^{(k-1)(j-1)}[b_1b_2 cdots b_j][a_1 a_2 cdots a_k]$$

provided the cycles $(a_1 a_2 cdots a_k)$ and $(b_1 b_2 cdots b_j)$ are disjoint.

The map $tilde Sigma_n to Sigma_n$ sends $[a_1 cdots a_k]$ to $(a_1 cdots a_k)$. So this gives you a corresponding presentation of the double of $A_n$ -- take your favourite presentation of $A_n$, lift the relators and see what happens using the above relations.

A small extra tidbit -- think of $Sigma_n$ as being the group of orientation-preserving isometries of $mathbb R^{n}$ that preserves a regular (n-1)-simplex. Then if you lift this group to $Spin(n)$, the extension you want is the one where $[a_1 a_2 cdots a_k]^k = -1$.

nt.number theory - Choosing a fast computer algebra system that works in characteristic p?

My personal experience is a few years old, but I don't think things have changed much. Sage is (or actually, was) more about ease of use then about performance. The only three CAS's you want to consider are

• Singular (Macaulay 2 uses Singular's engine)


• Cocoa.


• Magma.

Back then the fastest of the bunch was Magma, but not by much. Regarding ease of use, it was a tie between Macaulay 2 and Magma.

And now to some criticism: I never looked at Magma's code (proprietary), but I did look at both Singular and Cocoa. None of them uses SSE/GPGPU, which could probably give you an acceleration factor of 10-100.

at.algebraic topology - Characteristic classes in generalized cohomology theories?

Not a real answer to your question, but I think it may be related. One can ask the same question fofr Chern classes. For some generalized cohomology theories you can define Chern classes of complex vector bundles, and these satisfy the usual axioms for Chern classes.

What changes is the tensor product behviour. Given two line bundles L and M, the first Chern class $c_1(L otimes M)$ is given by a universal power series in $c_1(L)$ and $c_1(M)$. For instance in ordinary cohomology $c_1(L otimes M) = c_1(L) + c_1(M)$, while in K theory $c_1(L otimes M) = c_1(L) cdot c_1(M)$. Since line bundles form a group under tensor product, this power series is a (1-dimensional) formal group law.

So to any such cohomology theory you can attach a 1-dimensional formal group. For instance you attach the additive group to ordinary cohomology and the multiplicative group to K-theory. It turns out that you can go the other way round. The other 1-dimensional formal group laws are formal expansions of the group law of an elliptic curve near the origin; these give rise to the so called elliptic cohomology theories.

I think you can find details on the above constructions in any reference about elliptic cohomology. I've only heard about these theories, so I don't know an good reference. By the way, since I'm not an expert, please correct me if anything I have written above is wrong.

rt.representation theory - Definition of the symmetric algebra in arbitrary characteristic for graded vector spaces

What is the right definition of the symmetric algebra over a graded vector space V over a field k?

More generally: What is the right definition of the symmetric algebra over an object in a symmetric monoidal category (which is suitably (co-)complete)?

Two possible definitions come to my mind:

1) Take the tensor algebra over V and identify those tensors which differ only by an element of the symmetric group, i.e. take the coinvariants wrt. the symmetric group. The resulting algebra A is then the universal algebra together with a map V -> A such that the product of elements in V is commutative.

2) Take the tensor algebra over V and divide out the ideal generated by antisymmetric two-tensors. In this case, the resulting algebra A is the universal algebra together with a map V -> A such that the product of A vanishes on all antisymmetric two-tensors (one could say that all commutators of A vanish).

The definition 1) looks more natural and gives, for example, the polynomial ring in case V is of degree 0.

The definition 2) applied a vector space shifted by degree 1 gives (up to degree shift) the exterior algebra over the unshifted vector space. However, in characteristic 2 for example, one doesn't get the polynomial ring if one starts with a vector space of degree 0.

Finally, both definitions have a shortcoming in that they don't commute well with base change.