noncommutative algebra - When should I expect a quiver with potential to be rigid?

As I understand it, the quotient C(Q) = k[[Q]]/[,] is the (completed) vector space of formal linear combinations of oriented circuits in Q. Define a detour in Q to be an edge e and an oriented path p with the same source and sink as e. For every detour (e,p), there is a detour operator D(e,p). Given a circuit s, D(e,p)(s) is a sum of terms for each occurrence of e in s; each term is obtained by replacing that occurrence of e by p. Then the image I(S) of J(S) appears to be the span of all D(e,p)(S).

If I have all of that straight, then I can't think of a significant condition to guarantee that there is a rigid potential S. However, I can think of a significant condition to guarantee that there isn't one. Suppose for simplicity that the quiver Q has no parallel edges. Consider further just the shortest non-trivial circuits in C(Q); they span C(Q)_g, where g is the oriented girth of Q. For these circuits, the only detour operators D(e,p) that matter are the trivial ones with p = e. In other words, you have no choice but to replace an edge e with itself. But often C(Q)_g is bigger than the space of edges, i.e., Q could have more shortest cycles than it has edges. In this case, just for dimension reasons, I(S) can't contain C(Q)_g.

What bothers me is that if this is correct, then Derksen, Weyman, and Zelevinsky worked a little harder than necessary to make their Example 8.6.

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Now that I read on a bit more of the paper, the authors give some constructions of quivers with rigid potentials. For instance, they establish that the existence of a rigid potential is a mutation-invariant property of quivers. They also provide an example of a rigid quiver which does not mutate to an acyclic quiver. You can go a little further: In studying the cyclic part of the path algebra of a quiver, you can restrict attention to its strongly connected components. So you can start with any collection of strongly connected, rigid quivers, then connect them any way you like acyclically, and then mutate that.

It sounds like the status quo of the question is that there are several constructions of quivers that do not have rigid potentials, and several constructions of quivers that have rigid potentials, but that finding a good characterization of the dichotomy is an open problem.

Also, clearly the quivers that do not have rigid potentials are recursively enumerable. (I.e., there is an algorithm to confirm that a quiver does not have a rigid potential, but it might not terminate if it does have one.) Maybe a good remaining question is whether quivers with rigid potentials are also recursively enumerable, i.e., that the dichotomy is recursive.

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One way to confirm that a quiver has a rigid potential is to find a J(S) that contains all paths of some length. I(S) then automatically contains all loops of that length or greater. Moreover, it does not matter whether S includes any terms beyond the cutoff length. If a rigid potential always has this property, then determining whether there is one is recursive.

evolution - Is there any complex organism that is both autotroph and heterotroph?

Just look up kleptoplasty
http://en.wikipedia.org/wiki/Kleptoplasty
The slugs who most frequently employ it would fit your more restrictive definition. Now, if we're getting technical, they aren't born that way. They steal the chloroplasts, but you didn't say, so I guess you need to revise your expectations again. ;)

soft question - What happens to the boundary conditions as a PDE is approximated by a lesser order PDE?

The question seems to contain a misprint, as Harald Hanche-Olsen pointed out above. I am assuming that the real situation is as follows: $ epsilonnabla^4u + cnabla^2u-lambda u=F(x,y)$, where $ (x,y)inGamma $, with a pair of boundary conditions, say $ u=u'=0$, where $ (x,y)indeltaGamma $.

When $epsilon$ is "small" it is natural to assume that solutions of this PDE should be "close" in some sense to solutions of the reduced PDE $ cnabla^2u'-lambda u'=F(x,y) $.
This is a typical singular perturbation problem (see Van Dyke's "Perturbation methods in fluid mechanics" or any other text on asymptotic expansions that contains a discussion of singular perturbation problems). It is clear that since the order of the original equation has been reduced, one can now only satisfy one boundary condition, not two of the original problem. This basically indicates the presence of a boundary layer. This boundary layer is a solution of another asymptotic limit of the problem, the one for which $epsilonnabla^4u$ can no longer be omitted (see the comment by Willie Wong). The problem can generally be non-trivial in 2D; let me just show the idea how this type of problems is solved in 1D.

Consider for $xge 0$
$$ epsilon^2u_{,xxxx}-c^2u_{,xx}-lambda u=F(x) quadmbox{subject to}quad u(0)=u_{,x}(0)=0 $$
(I made some parameter changes to simplify the forthcoming algebra and changed the sign of $c^2$ as it actually makes more sense to pre-tension beam anyway). The "regular" solution $u_r$, as you pointed out, satisfies $c^2u_{r,xx}+lambda u_r=-F(x)$. The boundary layer solution is typically extracted by "zooming" into the vicinity of the boundary by using a coordinate transformation $xi=x/epsilon^{alpha}$ (so that $partial/partial x=epsilon^{-alpha}partial/partialxi$). How do we find $alpha$? Intuitively, we want to keep the previously vanishing term $epsilon^2u_{,xxxx}$ in the equation and the non-trivial answer should balance it with at least one other term of the original equation. If $alpha=1/2$ then first term is balanced by the third at $O(1)$, but the second term becomes $O(epsilon^{-1})$. Hence, one needs to take $alpha=1$ to balance the first two terms with the resulting equation
$$ epsilon^{-2}u_{,xixixixi}-c^2epsilon^{-2}u_{,xixi}-lambda u=F(x). $$
Clearly, when $epsilon$ is "small", the solution of this equation is approximated by the solution of boundary layer equation $u_{bl,xixi}-c^2u_{bl}=0$. What happens next depends on the sign of $c^2$. I shall take $c^2$ to be strictly positive. Then the boundary layer solution is a rapidly decaying exponent $u_{bl}=U_{bl}exp(-cxi)=U_{bl}exp(-cx/epsilon)$.

Let us now assume that the general solution of our problem is given by sum $u=u_r+u_{bl}$ (this is so-called Vishik-Lyusternik method). This means that when you were posing boundary condition $u_{,x}(0)=0$, you effectively posed
$$ u_{r,x}(0)+u_{bl,x}(0)=0 $$
Since $u_{bl}=U_{bl}exp(-cx/epsilon)$, then $u_{bl,x}=-cU_{bl}exp(-cx/epsilon)/epsilon=-cu_{bl}/epsilon$. Therefore, we can conclude that
$$ u_{r,x}(0)+u_{bl,x}(0)=u_{r,x}(0)-cu_{bl}(0)/epsilon=0 quadmbox{so that}quad u_{bl}(0)=epsilon u_{r,x}(0)/c. $$
If we now turn our attention to the original first boundary condition, the same line of argument gives
$$ u(0)=u_r(0)+u_{bl}(0)=u_r(0)+epsilon u_{r,x}(0)/c=0. $$

The conclusion then is: instead of solving the original fourth-order ODE with clamped boundary conditions, for sufficiently small $epsilon$ one can simply solve the following second order boundary value problem
$$ c^2u_{r,xx}+lambda u_r=-F(x) quadmbox{subject to the "effective" condition}quad u_r(0)+epsilon u_{r,x}(0)/c=0. $$
The resulting solution error will be of the order $O(epsilon^2)$ (except in the close vicinity of the boundary where we made error $O(epsilon)$). Your own suggested answer $u_r(0)=0$ is only accurate to within $O(epsilon)$ everywhere. My explanation is quite loose here, but I hope it should give you the idea of how to tackle this kind of problems.

dna - Should the length of the electrodes in the electrophoresis chamber be proportional to chamber's size?

I am trying to build a small horizontal electrophoresis chamber from scratch. I want to use it for comet assay and I will be using only 1 slide, so it's going to be about 3cm wide, 10cm long and 4cm high, just enough to fit the slide and a small amount of buffer. As far as I know such thing is not commercially available, so before I make it I want to make sure that it's going to work. The main issue is the cost of the electrode (has to be platinum wire). I don't want to spent that much money just to realize that it doesn't work.
My question is: is a 3cm electrode in a 60-70cm³ enough? Basically, are there any physical restrictions on the size and the position of the electrodes and the shape of the chamber?
Since I am a biologist and my knowledge in physics are quite limited, I would greatly appreciate any information about this.

Attaching suspension cells to coverslip during mycoplasma contamination detection

I am planing to screen my cell cultures for mycoplasma contamination using the Hoechst 33258 DNA staining method.

This method is suitable for adherent cultures. Is this method suitable for suspension cultures?

If yes, what is the procedure to attach my suspension culture on coverslip and prevent washout of cells during staining.

ac.commutative algebra - Examples of finite local rings of length 2 or 3

I'm just going to consider the local rings with residue field $mathbf{F}_p$.

Suppose A' is an extension of $mathbf{Z}/p$ with ideal $mathbf{Z}/p$. If we take the fiber product with $mathbf{Z}$, we get an extension of $mathbf{Z}$ with ideal $mathbf{Z}/p$. Up to isomorphism, there is only one such extension: $B' = mathbf{Z}[t] / (t^2, pt)$. The kernel of the map from B' to A' is isomorphic to $mathbf{Z}$ and reduces modulo t to the ideal generated by p. Therefore, the square-zero extensions of $mathbf{Z}/p$ are all isomorphic to $mathbf{Z}[t] / (t^2, pt, p + lambda t)$ for some $lambda not= 0$. If $lambda$ is not a multiple of p, we get $mathbf{Z} / p^2$; if $lambda$ is a multiple of $p$, we get $mathbf{Z}[t] / (p, tp, t^2) = mathbf{F}_p[t] / t^2$. So these are all the length 2 finite local rings with residue field $mathbf{F}_p$.

For length 3, we'll look for extensions of $mathbf{Z} / p^2$ by $mathbf{Z} / p$. The same analysis shows that these are all of the form $mathbf{Z}[t] / (t^2, pt, p^2 + lambda t)$. If $lambda$ is not divisible by $p$, we get $mathbf{Z} / p^3$ and if $lambda$ is divisible by $p$ we get $mathbf{Z}[t] / (p^2, pt, t^2)$.

We also have to look for extensions of $mathbf{F}_p[t] / t^2$. By base change, any such extension A' gives an extension B' of $mathbf{Z}[t]$ with ideal $mathbf{Z} / p$. Once again, there is only one of these up to isomorphism (since $mathbf{Z}[t]$ is projective, for example) and it is given by $mathbf{Z}[t,u] / (u^2, pu, tu)$. The ideal of the map from B' to A' generates the ideal of the map from $mathbf{Z}[t]$ to $mathbf{F}_p[t] / t^2$. Since this is generated by p and t^2 the ideal of A' in B' is generated by $(p + lambda u, t^2 + mu u)$ and A' is of the form

$mathbf{Z}[t,u] / (u^2, pu, tu, p + lambda u, t^2 + mu u)$

for some polynomials $lambda, mu in mathbf{Z}[t]$.

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Edit: If $lambda$ is not in $(p, t)$ then it is invertible in the quotient, so we get

$mathbf{Z}[t,u] / (p^2, tp, t^2 + mu lambda^{-1} p)$.

There are two possibilities up to isomorphism here, depending on whether $- mu lambda^{-1}$ is a quadratic residue modulo p.

If $lambda$ is in $(p,t)$ we get

$mathbf{Z}[t,u] / (u^2, pu, tu, p, t^2 + mu u) = mathbf{F}_p[t,u] / (u^2, tu, t^2 + mu u)$.

The $mu$ is also not in $(p, t)$ then we get $mathbf{F}[t] / t^3$. If $mu$ is in $(p,t)$ we get $mathbf{F}_p[t,u] / (u^2, tu, t^2)$.

Modulo any mistake I made above, I think a complete list is of length 3 finite local rings with residue field $mathbf{F}_p$ is

1. $mathbf{Z} / p^3$,


2. $mathbf{Z}[t] / (p^2, pt, t^2)$,


3. $mathbf{Z}[t] / (t^2 - p, t^3)$,


4. $mathbf{Z}[t] / (t^2 - alpha p, t^3)$ where $alpha$ is a non-quadratic residue modulo $p$,


5. $mathbf{F}_p[t] / t^3$, and


6. $mathbf{F}_p[t,u] / (t,u)^2$.

dg.differential geometry - Changing coordinates so that one Riemannian metric matches another, up to second derivatives

Let $g$ and $g'$ be two $C^2$-smooth Riemannian metrics defined on neighborhoods $U$ and $U'$ of $0$ in $mathbb R^2$, respectively. Suppose furthermore that the scalar curvature at the origin is $K$ under both metrics.

My question: Is there a coordinate transformation taking one metric to the other, such that they agree up to second derivatives at the origin? i.e., if $x : U to U'$ is the transformation, we have

$g_{ij}' = g_{ab} ~x_i^a x_j^b$,

evaluating everything at $0$; there are similar equations for the first and second derivatives. Clearly this is false if the scalar curvatures aren't equal. I don't care what happens away from the origin.

In the excellent thread When is a Riemannian metric equivalent to the flat metric on $mathbb R^n$?, Greg Kuperberg says:

If remember correctly, there is a more general result due to somebody, that any two Riemannian manifolds are locally isometric if and only if their curvature tensors are locally the "same".

If "local isometry" means that the metrics are equal on a neighborhood of the origin, then the metrics I have in mind are not locally isometric, since the only information I have is that their curvatures match at one point.

Edit: I'm pretty sure that Deane answered my question, but let me clarify. Let $g_{ij}$ be some "reasonable" metric, e.g. a bump surface metric, and consider a point $p$ where the scalar curvature is $K$. Let $g_{ij}'$ be an arbitrary metric on a neighborhood $U$ of the origin in $mathbb R^2,$ with scalar curvature $K$ at $0$.

Then the question becomes: does there exist a coordinate change on the bump surface such that the equation $g_{ij}'(0) = g_{ab}(p) ~x_i^a x_j^b$ is satisfied, as well as the corresponding equations for the first and second derivatives? That is, there are $18$ pieces of pertinent information

$(*)~~~g_{11}', g_{12}', g_{22}'; g_{11,1}', g_{12,1}', g_{22,1}', g_{11,2}', g_{12,2}', g_{22,2}'; g_{11,11}', g_{12,11}', g_{22,11}', g_{11,12}', g_{12,12}', g_{22,12}', g_{11,22}', g_{12,22}', g_{22,22}'$.

I want to change coordinates on my nice surface such that the metric and its derivatives line up with $(*)$.