User johannes hahn - MathOverflow

Answer by Johannes Hahn for Spectrum and scheme of the commutative group-algebra of an abelian group.

2013-02-20T13:15:20Z

Hi.

Varying the coefficients gives certainly a lot of information about the group. For example the smallest field $K\supseteq\mathbb{Q}$ such that $K[G]$ becomes split semisimple (which means isomorphic to $K^G$ in this case) encodes the exponent of the group (which also can be read of from the Loewy length of the modular group algebras I think).

If you are willing to consider the scheme including the involution $\ast: R[G]\to R[G]$ which is defined by $g\mapsto g^{-1}$, then the group is in fact determined up to isomorphism by $\mathbb{Z}[G]$ since $\lbrace\pm 1\rbrace G$ is the group of "orthogonal" units: $\lbrace x\in\mathbb{Z}[G] \mid xx^\ast=1\rbrace = \lbrace\pm1\rbrace G$.

Answer by Johannes Hahn for Irreducibility of Coxeter Graphs as a Function of Generating Sets

2013-01-18T03:28:25Z

Yes, for $S_n$ the answer is obvious: None of the $S_n$ is decomposable hence there cannot be a reducible Coxeter system for $S_n$.

In general one can do the following construction: $W/W' = C_2^k$ where $k$ is the number of connected components in the graph that is obtained from the Coxeter graph of $(W,S)$ by deleting all edges $s-t$ with $ord(st)\in 2\mathbb{N}\cup\lbrace\infty\rbrace$. In particular $W/W' = C_2$ for all finite Coxeter groups other than $B_n$ and $I_2(m)$ with even $m$. If there was a decomposition $W \cong W(X_1 \times X_2)$ for two nonempty Coxeter diagrams $X_1$ and $X_2$, then we would have $\dim_{\mathbb{F}_2} W/W' \geq 2$. This is a contradiction. Therefore we only need to care about the types $B_n$ and $I_2(m)$ with $m$ even.

If $m$ is even, then $I_2(m)$ is isomorphic to the Coxeter group of $A_1\times I_2(\frac{m}{2})$.

For $W(B_n) = (\mathbb{Z}/2)^n \rtimes S_n$ with uneven $n\geq 3$ the answer is also positive because the natural $\mathbb{F}_2[S_n]$-module $\mathbb{F}_2^n$ is decomposable: $C_2^n \rtimes S_n = X \times (Y \rtimes S_n)$ with $X=\langle(1,1,...)\rangle$ and $Y:=\lbrace x | \sum_i^n x_i = 0\rbrace$. This decomposition is a Coxeter group of type $A_1\times D_n$.

I think the answer should be negative for $W(B_n)$ with even $n\geq 4$ but I cannot think of a short argument at the moment. Here is what I got so far:

If $W(B_n) = N_1\times N_2$ then the projections of $N_i$ must be commuting, normal subgroups of $S_n$ that generate $S_n$ together. Ergo one of them projects onto 1 the other on $S_n$. Therefore we may assume wlog $N_1\leq (\mathbb{Z}/2)^n$ and for every $\sigma\in S_n$ there is a $(v_\sigma, \sigma)\in N_2$. Since $N_1$ is abelian, it acts trivially on itself. $N_2$ acts trivially on $N_1$ by assumption, hence $N_1\subseteq Z(W)=\langle(1,1,...)\rangle=X$.

The length-function of $B_n$ maps $(1,1,...)$ to zero since $n$ is even. The length-function is a non-zero homomorphism $N_1/N_1' \times N_2/N_2' = W/W' \to \mathbb{Z}/2$ and it maps $N_1$ to zero. Because we know $W/W' \cong (\mathbb{Z}/2)^2$ we get $\ker(l)/W' = N_1/N_1'$ and therefore $\ker(l) = N_1 \times N_2'$. Hence $v:=(1,1,0,...)$ lies in $N_1\times N_2'$. Therefore either $v$ itself or $v+(1,1,1,...) = (0,0,1,...)$ lies in $N_2$. The conjugates of both vectors generate $Y:=\lbrace x | \sum_i v_i = 0\rbrace$. Hence $N_1=X\leq Y\leq N_2$ and we finally have a contradiction: $W(B_n)$ is indecomposable if $n$ is even.

EDIT: I totally forgot $F_4$. Well one can show that $W(D_4)$ is indecomposable as a group similar to the above proof: $W:=W(F_4) = W(D_4) \rtimes S_3$ where $S_3$ acts as the group of diagram automorphisms. In particular $s_1, s_2, s_3 s_2 s_3, s_4 s_3 s_2 s_3 s_4$ are a set of simple reflections for the $D_4$ subgroup and $s_3, s_4$ generate the $S_3$ subgroup.

If $N_1, N_2$ are direct factors of $W$, then wlog $N_2 \twoheadrightarrow S_3$ and $N_1\subseteq C_{N_2}(W) = \langle s_2, s_3 s_2 s_3, s_4 s_3 s_2 s_3 s_4 \rangle \cong C_2^3$ is abelian and hence central. But now we have $N_1\subseteq Z(W) \cap W(D_4) \subseteq Z(W(D_4)) = 1$.

Answer by Johannes Hahn for commutative rigs and the Grothendieck Group

2012-08-30T20:36:02Z

The Grothendieck group of a commutative group is group itself, hence applying the construction twice doesn't change the result. The semiring case is a special case of this.
What do you mean by "order of application". There is no order if you apply the same thing twice. G(G(M)) is the same as G(G(M)) ...
The Grothendieck group of the naturals is the ring of integers. This is the standard construction of the integers.

This sounds somewhat to trivial ... May be you mean something different here. Maybe you want to apply the construction to the multiplicative monoid instead of the additive structure in one of the two steps? The result in this case is (independent of the order) the zero ring because there is an absorbing element in the multiplicative monoid (the zero element of the semiring) so that all elements of the monoid get identified in the Grothendieck group.

Answer by Johannes Hahn for Can we actually find any fixed points with Brouwer's theorem?

2012-08-29T21:43:30Z

There is a constructive version of Brouwer's theorem via Sperner's theorem. This gives an actual way to compute fixed points. (Not a very efficient one, granted, but an algorithm nonetheless)

Answer by Johannes Hahn for Manifold-Valued Sobolev Spaces

2012-07-12T07:31:58Z

Hi.

This isn't well-defined (even for higher order smoothness). Even the case $M=\mathbb{R}$ doesn't work, because arbitrary diffeomorphism can have arbitrarly bad growth towards infinity and therefore do not map H^1 map to H^1 maps.

Answer by Johannes Hahn for ODE for functions with values in locally convex TVS

2012-05-02T14:21:41Z

Hi.

By using functionals one can reduces uniqueness to the one-dimensional case: If $u$ is a solution to $\frac{d}{d}t u = f(t,u)$ then $\lambda u$ is a solution of $\frac{d}{dt} (\lambda u) = (\lambda f)(t,u)$ for all $\lambda\in V'$. In your case all the functions $\lambda f$ satisfy a Lipschitz-condition so Picard-Lindelöf gives you the uniqueness of $\lambda u$. Since the values $\lambda u$ determine $u$ uniquely in a LCS, this means $u$ is unique. A lot a other problems can be handled completely analogously.

The more difficult question is the existence of $u$. I'm not sure, but I think one could use Schauder's fixed-point theorem after one has replace the ODE with an integral equation. (Using Pettis-integrals for example).

Answer by Johannes Hahn for Is there a polynomial equation whose solution over the integers is independent of ZFC

2012-04-28T13:49:45Z

Hi Daniel.

The point with Hilbert's 10th problem is that diophantine equations are complex enough to encode Turing machines and other complicated stuff that has some kind of no-can-do-theorem. In particular: To each Turing machine there is a polynomial $p\in\mathbb{Z}[X_1,\ldots,X_n]$ such that $p$ has a solution in $\mathbb{N}$ iff the machine halts (lets consider only turing machines with empty input for simplicity).

Now lets look at the machine that lists all PA-proofs and halts iff it finds a proof of a given PA-statement $\psi$. If PA could prove the existence or non-existence of solutions to every diophantine equation, then it could decide this in particular for those equations that encode Turing machines, i.e. this algorithm would effectively decide the halting problem which is impossible.

There are other things that you can encode with diophantine equations. For example it is possible to translate a statement like $con(PA)$ into a diophantine equation. Now PA cannot prove the existence of solutions of such an equation because that would mean that PA proves $con(PA)$ which is also impossible. It can't disprove the existence of solutions either because then PA would prove $\neg con(PA)$ which is also impossible because PA is consistent.

Now about the models... There isn't that much to say about it. If PA cannot decide the existence of solutions of $p=0$, then the standard model $\mathbb{N}$ won't contain a solution (because if it would, this solution could be explicitely written down and it could be checked by calculation that it is indeed a solution thus giving a formal PA-proof of its existence). But there will be many non-standard models with solutions. Those solutions are of course non-standard numbers, so you can't write them down or do any kind of explicit computations with them. In particular: You won't be able to squeeze an existence proof out of them like you can with a standard solution.

I hope that answers your questions.

Are there Hamilton paths in Cayley graphs of Coxeter groups?

2012-03-20T16:35:20Z

Hi everyone.

I want to optimize certain computation on finite Coxeter groups $(W,S)$. Basically I compute the matrices $\rho(T_w)$ for all $w\in W$ of a matrix representation $H\to K^{d\times d}$ of the Hecke algebra $H=\mathcal{H}(W,S)$ and do some stuff with these matrices. The representation is given as a list of the matrices $\rho(T_s)$ for $s\in S$. The obvious way to do such a computation is to use the property $l(ws)=l(w)+1 \implies T_{ws}=T_w T_s$ of the standard basis of $H$ to move from layer to layer in the group ("layer" meaning sets of the form $\lbrace w\in W | l(w)=k\rbrace$ for fixed $k$) and by multiplying the matrices $\rho(T_s)$ to the existing ones.

Since I'm also interested in big examples, I quickly run into trouble with my memory in this way because to compute the $\rho(T_w)$ with $l(w)=l$ one has to store all the $\rho(T_y)$ with $l(y)=l-1$ which can be quite a big number if $l$ is around $\frac{1}{2}l_{max}$. Even though I have access to a machine with 128GB RAM, this is too much if $W$ and the dimension of $\rho$ are big.

A few days ago I read about Hamilton paths in Cayley graphs. This would solve my memory problem, because if I knew a Hamilton path $w_1,\ldots,w_n$ I would only need to store the single matrix $\rho(T_{w_i})$ to compute $\rho(T_{w_{i+1}})$ and forget about it afterwards. If I had access to a Hamilton path in the Cayley graph $\Gamma(W,S)$ I could carry out my calculations with using only little more memory than I already need for the input itself.

Googling showed my that in general it is not even clear if such hamilton paths always exists. That's rather unfortunate, but on the positive side I also found out that there is an easy algorithm in case of the symmetric group and its Coxeter generating set. So I'm hoping that there is a result in the case of Coxeter groups.

So my questions are:

If $(W,S)$ is a finite Coxeter system, does there exists a Hamilton path in the Cayley graph $\Gamma(W,S)$?
If this is indeed the case, is there an easy algorithm to traverse a Hamilton path?

Efficient enumeration of Bruhat intervals

2011-11-03T21:11:58Z

Hi everyone.

I'm currently programming some stuff for Hecke algebras. My current implementations have several bottlenecks and I'd like to improve that as much as I can so that I can use stuff like $E_8$ as input one day (without waiting until the heat death of the universe for an answer).

Given a Coxeter group W (let's start with a finite one), one of the things that I would like to improve are some iterations over intervals in the Bruhat order in W. For example a recursion for the Kazhdan-Lusztig-polynomials $P_{y,w}$ and the $\mu$-polynomials $\mu_{y,w}^s$ involves formulas like
$P_{y,w}^\ast = v_t P_{y,w}^\ast + P_{ty,tw}^\ast - \sum_{z\in W: y\leq z < tw \wedge tz < z} P_{y,z}^\ast \mu_{z,tw}^t $
The crucial step of calculating $\mu_{y,w}^s$ involves a similar sum over the interval $(y,w)$. Some other calculations I'd like to perform in the future also have this structure.

One way (this is how I do it at the moment) of realizing this summation over intervals $[a,b]$ is to iterate through all elements of $W$ whose length is between $l(a)$ and $l(b)$ and check everytime if $a\leq z \leq b$ is satisfied. This works fine if both $a$ and $b$ are near the bottom or near the upper end of the Bruhat order, but if they are in the middle, the level sets $\lbrace z | l(z)=const\rbrace$ are much, much bigger than the interval $[y,z]$: Considering the characterization of the Bruhat order in terms of subwords something like $2^{l(w)-l(y)}$ should be an upper bound for the cardinality of this interval, but the level sets in the middle of the Bruhat order have more than $|W|/l(w_0)$ (which is more than $5,8*10^6$ in the $E_8$ case).

A solution could be to use the subword-characterization and iterate through all subwords of $w$. It is no difficult to write a program iterates that way through $[1,w]$. But this approach enumerates all $2^{l(w)}$ subwords and discards several of them to ensure that only reduced words appear and each appears only once. Now an additional check whether $y\leq z$ or not only slows thing down even further. So this also is not a very efficient way if $w$ is somewhere in the middle of the Bruhat order.

Hence my question is:
Is there an efficient way to enumerate all elements of Bruhat intervals?

Answer by Johannes Hahn for Is there a formula for the size of Symplectic group defined over a ring $Z/p^k Z$?

2012-02-08T19:43:13Z

Hi.

EDIT: As Joe Silverman pointed out, this approach doesn't work as simple as I imagined it. Sorry for that. I leave the attempted proof here in case someone has an idea how to fix it.

Yes, there is such a formula. It works for in a similar fashion all of the classical algebraic group. Consider the projection $\mathbb{Z}/p^k\to\mathbb{Z}/p$. It induces a surjection $Sp_{2n}(\mathbb{Z}/p^k)\to Sp_{2n}(\mathbb{Z}/p)$ because every transvection has a preimage and the transvections generate $Sp_{2n}(F)$ for every field $F$. The kernel of the homomorphism consists obviously of those matrices $M$ with $M-I \in (p\mathbb{Z}/p^k\mathbb{Z})^{n\times n}$. There are $p^{(k-1)n^2}$ such matrices.

Therefore $|Sp_{2n}(\mathbb{Z}/p^k)| = p^{(k-1)n^2} |Sp_{2n}(\mathbb{Z}/p)|$. The order of the symplectic group over $\mathbb{Z}/p$ is known.

Sequential topological vector spaces

2010-08-21T14:18:37Z

Since I'm dealing with the distinction between sequential continuous and continuous maps at the moment I came to ask myself once again what can be said about spaces where these two notions agree (sequential spaces). Of course we all know that metric spaces and more generally first-countable spaces are sequential and in the literatur it seems that often metrizability or first-countability is only assumed in order to not need to distinguish between sequential continuity and continuity.

I'm mostly interested in spaces that arise naturally in functional analysis, i.e. subspaces of topological vector spaces. A well known theorem says that a hausdorff topological vector space is metrizable iff it is first-countable. I tried to find out what could be said about sequential t.v.s. Are sequential t.v.s. metrizable too? Are there any reasonable t.v.s. that are sequential but not metrizable?

Answer by Johannes Hahn for endomorphism rings of indecomposable objects

2011-07-22T17:29:51Z

Hi benjamin.

You're asking if $\mathcal{C}$ is a so called Krull-Schmidt-category. There are several sufficient conditions know from ring theory if $\mathcal{C}$ is a module category. For example the argument George mentioned gives an affirmative answer if $\mathcal{C}$ is the category of finitely generated (left)modules over an artinian ring. You get another characterisation by considering semiperfect rings. A ring is semiperfect iff the category of finitely generated projective (left)modules is Krull-Schmidt. Examples of semiperfect rings include all left or right artinian ring as well as all finitely generated $R$-algebras where $R$ is a complete local ring or a discrete valuation ring.

This connection between semiperfect rings and Krull-Schmidt-categories is in fact somewhat stronger: One can show that $\mathcal{C}$ is Krull-Schmidt iff $End_\mathcal{C}(X)$ is semiperfect for all objects $X\in\mathcal{C}$.

Answer by Johannes Hahn for Algebra with positive definite symmetrizing trace is semisimple.

2011-06-29T12:47:27Z

Though not an answer to the request for references, I'd like to add the fact that these algebras are trivial:

If $A$ is a split, symmetric $\mathbb{R}$-algebra with positive definite trace form (in the above mentioned sense), it is semisimple as proven above. It is therefore a product of matrixrings over $\mathbb{R}$ and the traceform is a linear combination with positive coefficients of the traces of these matrixrings. But the trace on $\mathbb{R}^{d\times d}$ is not positive definite if $d>1$ because there are nonzero matrizes with square zero. Therefore we get an algebra isomorphism $A \cong \mathbb{R} \times \ldots \times \mathbb{R}$.

If there is indeed a generalization of this result to other fields than $\mathbb{R}$, this proof would probably carry over.

Answer by Johannes Hahn for When does a symmetric algebra over a field of characteristic 0 fail to be semisimple?

2011-06-27T11:30:15Z

The answer is no. The reason is that every algebra can be embedded into a symmetric algebra, the so called trivial extension:

If $A$ is a $K$-algebra, then define $D(A):=A\oplus Hom_K(A,K)$. $I:=Hom_K(A,K)$ is a $A$-$A$-bimodule via

$a\cdot \phi \cdot b:=x\mapsto \phi(bxa)$

Hence you can define an $K$-algebra structure on $D(A)$ such that $I$ becomes an ideal with $I^2=0$. The multiplication is explicitly given by:

$(a+\phi)(b+\psi) := ab+a\cdot\psi+\phi\cdot b$

The trace form is given by

$(a+\phi,b+\psi) := \psi(a)+\phi(b)$

Now $I$ is a nilpotent ideal and therefore contained in the Jacobsen radical of $D(A)$. In particular do $A$ and $D(A)$ have the same simple modules and $D(A)$ is split iff $A$ is. But because $J(D(A))\neq 0$ the $K$-algebra $D(A)$ is never semisimple regardless of what field $K$ you started with.

Answer by Johannes Hahn for Examples of Amenable Groups other than Z_n

2011-04-19T20:18:44Z

All compact hausdorff groups are amenable because they have a Haar-measure which can be chosen with $\mu(G)=1$. So $O_n(\mathbb{R})$, $U_n(\mathbb{C})$, closed subgroups thereof as well as quotients by closed normal subgroups are amenable.

Answer by Johannes Hahn for What are some examples of colorful language in serious mathematics papers?

2011-04-07T10:02:47Z

In T.Y.Lams book "Lectures on modules and rings" there is a chapter on quotient rings. The three subsections of which are named "The Good", "The Bad" and - of course - "The Ugly". The three subsections are about existence and uniqueness of a "localization" with the universal property in the noncommutative case ("The Good" though nothing is good about this localization in general, everything nice is lost in the general case), Mal'cev's example of a domain that cannot be embedded into a division ring ("The Bad") and further theorems about which classes of rings can be embedded together with example that there need not to be a unique minimal such division ring ("The Ugly").

Answer by Johannes Hahn for Why don't ideals and quotients work well for categories?

2011-04-01T15:49:17Z

You're saying, that the definition of an Ideal $I(X,Y)\leq Hom(X,Y)$ does not affect objects. That is not true, because in the quotient category there can be more objects isomorphic to one another than before which is exactly the same behaviour in groups and ring where elements can become identical by passing to a quotient.

Consider this: If C is the category of finitely generated modules over an artinian ring (more general: modules of finite length over an aribitrary ring) and A is a full, isomorphism closed subcategory that is also closed w.r.t. taking direct summands, then define the following ideal: $I_A(X,Y):=\lbrace f:X\to Y | \exists P\in A: f \text{factors through} P\rbrace$

Then one can show that in the quotient category $C/A:=C/I_A$ the following charaterizations hold:

$X\cong^{C/A} 0 \iff X\in A$ and more general

$X\cong^{C/A} Y \iff \exists P_0, Q_0\in A: X=X_0\oplus P_0, Y=Y_0\oplus Q_0, X_0 \cong^C Y_0$ (i.e. "isomorphic in C/A" means nothing else than "isomorphic in C if direct summands from A are neglected".

This construction with A:={projective modules} gives you the stable module category of a ring. If you're dealing with group rings then things like the Green-correspondence become equivalences of such quotient categories where certain subcategories of relative projective modules are chosen as A.

Answer by Johannes Hahn for Problems where we can't make a canonical choice, solved by looking at all choices at once

2010-12-21T13:19:59Z

Another classical example of "looking at all choices instead of one" is the idea of the fundamental groupoid of a topological space. Instead of choosing one base point and letting all loops begin and end in this point one considers all paths between all points (modulo homotpy). This notion makes theorems like the Seifert-Van-Kampen-theorem much more natural. One does no longer have to add technical conditions that certain intersections contain the base point and are path connected.

Answer by Johannes Hahn for Weak and Strong Integration of vector-valued functions

2010-12-04T23:25:55Z

Hi.

One issue with Bochner integration is that it does not include Riemann-integration. There are Banach-space-valued R-integrable functions that are not B-integrable (example: Consider $X:=\mathcal{l}^p([0,1])$ with $2\leq p < \infty$ and $f:[0,1]\to X, f(t):=e_t$ where $e_t$ is the tupel with exactly one equal to 1 and all other components equal to 0.). The Gelfand-Pettis-Integral on the other hand includes both the Bochner- and the Riemann-Integral.

One problem with B-integration is that you need functions that are almost separable valued (meaning: there is a nullset whose complement has separable image) in order to approximate them with simple functions and this may be a strong restriction. Another issue is that for certain applications the weak topologies just behave better than the strong ones so that Pettis-integration is the natural notion of integration in this cases.

Answer by Johannes Hahn for Generalized notions of solutions in various areas of mathematics

2010-10-07T13:25:12Z

Well, then I'll start with the most obvious generalized solutions:

weak solutions to PDEs
Schwartz's generalized Functions aka Distributions,
Colombeau's algebra(s) of generalized functions and
various other kinds of generalized functions
Quasi-Minima in functional analysis: A quasi-minimum of a functional $\mathcal{F}$ is a $u$ such that $\mathcal{F}u\leq Q\mathcal{F}v$ for all $v$ (with some constant $Q\geq 1$)
Every solution of an polynomial equation within $\mathbb{C}$ can be a generalized solution if you're problem is something that has only real (maybe some geometric problem) or only integer or even only natural (maybe something from number theory) solutions. But considering all complex solution to your particular equation often gives a very elegant treatment of the problem.

Answer by Johannes Hahn for Moser iteration for elliptic systems

2010-09-15T09:39:19Z

Hi.

The point is not the ellipticity. In fact the Argument of De Giorgi, Moser and Nash was designed for elliptic problems. The point is that solutions $u: \Omega\to\mathbb{R}^N$ of elliptic problems with $N>1$ just aren't $C^{1,\alpha}$ any more in general. This is no problem with the method, it's intrinsic. The famous counterexample is by De Giorgi himself.

See Giusti - The direct method of variational calculus for more details to this topic. The counterexample itself can be found as example 6.3 in this book. The paper from De Giorgi is "Un esempio di estremal discontinue per un problema variazionale di tipo ellittico", Boll. U.M.I., 4 (1968), 135-137

Answer by Johannes Hahn for Conditions on a metric space so that boundedness implies total boundedness

2010-08-27T15:55:32Z

If $X$ is locally compact, then it has this Heine-Borel-property. For topological vector spaces locally compactness is equivalent to finite dimension if I remember correctly.

But there are other examples even vector spaces that have the Heine-Borel-property without being locally compact. The space $H(U)$ of holomorphic functions on an open set $U\subseteq\mathbb{C}$ with the topology of locally uniform convergence of all derivatives. This is Montel's theorem and therefore such spaces are called Montel spaces (well a certain additional condition is needed, but that's not the point) Another example is the Schwartz-Space $\mathcal{S}(\mathbb{R}^n)$ of rapidly decreasing functions. Because being Montel is stable under taking strong duals, the space of tempered distributions $\mathcal{S}'(\mathbb{R}^n)$ has the Heine-Borel-property too (but is not metrizable).

Answer by Johannes Hahn for Are there alternative proofs for existence/uniqueness of ODE solutions?

2010-08-26T18:49:18Z

A very interesting kind of existence (though not uniqueness) proofs are proofs that use one of the various fixed point theorems and the tools from fixed point theory: The Schauder fixed point theorem can be used to prove Peano's existence theorem or simple existence theorems for boundary value problems. The theory of Brouwer-degree of certain mappings (not just between manifolds but also between banach spaces) can be used to prove several existence theorems, for example existence of periodic solutions for certain ODEs.

Granas / Dugundji - Fixed point theory is a very good (and very densely written) book about all kinds of fixed point theorems and one of the main application that frequently occurs throughout the book are existence theorems for differential equations.

Then there is the variational approach: Finding minima of functionals often is the same thing as finding solutions of PDEs (Euler-Lagrange-equations !). So very much of functional analysis can be applied to prove various existence theorems for PDEs. For example have a look at Guisti - Direct methods in the calculus of variations. Any book on the finite element method (Braess comes to mind, but I'm not sure at the moment if it is in english or in german... may be there are two versions?) will show you how Hilbert space methods can be applied to prove existence theorems.

The whole theory of distributions was invented for dealing with (linear) PDEs, their (weak) solutions and the regularity theory of these solutions. The Ehrenpreis-Malgrange-theorem is a very strong existence theorem that says that all linear PDEs with constant coefficient have distributional solutions. In Fact there exist (tempered) Green's functions for every such PDE.

And there is of course a more heavy machinery too: Morse theory and generalizations of it were used for (invented for?) the proof of the Arnol'd conjecture which also shows the existence of certain periodic solutions of differential equations.

Answer by Johannes Hahn for Graph properties, categorically defined

2010-08-26T09:56:07Z

There are some subtle point concerning the definition of this category. It has different properties if you define "graph" in a different way.

If you allow your graphs to have loops, then the category of all graph becomes a topological category over SET. In particular all (small) limits and colimits exist in this category. Sometimes it is best to consider only the graphs with loops at all vertices (if you just don't draw them this subcategory can be identified with the category of all simple graphs). The product in this category is the tensor product of graphs. A connected object (in a topological category an object is connected if all morphisms into discrete objects are constant) in this category is exactly a connected graph.

The subcategory of graphs without loops is not that well behaved from the point of view of category theory, But graphs without loops have their merits too: A graph morphism from $(V,E)$ to the complete graph without loops on $n$ vertices is the same thing as a $n$-vertex-coloring of $(V,E)$.

Answer by Johannes Hahn for Interesting applications (in pure mathematics) of first-year calculus

2010-08-10T16:41:42Z

The intermediate value theorem is a basic ingredient in a Galois theory-based proof of the fundamental theorem of algebra. It is used as "Every real polynomial of odd degree has a real zero".

Answer by Johannes Hahn for List of centers of finite groups

2010-06-26T20:49:47Z

One can calculate the center of a semidirect product $U\ltimes V$ very explicit. Let $\phi: V\to Aut(U)$ the corresponding homomorphism. Then the following holds: $$(u,v)\in Z(U\ltimes V) \iff v\in Z(V), u\in C_U(V), \kappa_u=\phi(v^{-1})$$ where $\kappa_u$ is the conjugation with $u$. In particular $\phi(v)\in Inn(U)$. If $U$ is abelian, this gives $Z(U\ltimes V)=Z(V) \cap \ker(\phi)$. If $V$ is abelian too, then $Z(U\ltimes V)=\ker(\phi)$. Now all examples you're looking for can be easily written down.

Answer by Johannes Hahn for Difference between represented and singular cohomology?

2010-06-24T20:42:50Z

Singular cohomology and the definition via Eilenberg-MacLane-spaces give the same cohomology theory. There no examples where they differ, $H^n(X,G)$ and $[X,K(G,n)]$ are naturally isomorphic.

If you're looking for an example of two essentially different cohomology theories with the same coefficient group and both satisfying the dimension axiom, then Cech-cohomology is what you're after. The topologist's sine surve is a space with $H^0(X;\mathbb{Z})=\mathbb{Z}^2$ and $\check{H}^0(X;\mathbb{Z})=\mathbb{Z}$.

Answer by Johannes Hahn for Why free topological groups on Tychonoff spaces?

2010-05-08T17:44:18Z

Hi.

This is mainly because one wants to have $X$ as a subspace of $F(X)$. Since every topological group is tychonoff (:=closed sets can be seperated from points outside by continouos functions) and so is every subspace. So being Tychonoff is necessary (and sufficient) for $X$ being a subspace of $F(X)$.

Answer by Johannes Hahn for Is it true that the only interesting topologies are metric topologies and weak topologies?

2010-04-16T14:02:20Z

Frechet spaces, limits of Frechet spaces were mentioned before. I'd like to emphasize a particularly important example (which was also mentioned before, but I want to extend it a bit): The space of test functions $\mathcal{D}(\Omega)$ is a strong inductive limit of Frechet spaces, neither metrizable nor do they carry the weak topology. These spaces are the foundation of distribution theory and therefore most important.

on the other hand the other two usual spaces of test function $\mathcal{E}(\Omega)$ and the schwartz space $\mathcal{S}(\mathbb{R}^n)$ are metrizable because they can be topologized by a countable family of (semi)norms.

The distribution space $\mathcal{D}'(\Omega), \mathcal{E}'(\Omega), \mathcal{S}'(\mathbb{R}^n)$ can be endowed with the weak topology (that is the pointwise convergence). But the strong topology is also common and this is again neither weak nor metrizable.

Answer by Johannes Hahn for If the fourier transformed of f is odd, is f odd?

2010-03-31T20:33:48Z

Look at $L^2$ first: In $L^2$ the FT is diagonalizable. The space of odd functions $\in L^2$ is the direct sum `$Eig(\mathcal{F},+i)\oplus Eig(\mathcal{F},-i)$ of the eigenspaces of $\mathcal{F}$ with respect to the eigenvalues $+i$ and $-i$. Because eigenspace are mapped into themselves, $\mathcal{F}$ maps odd functions to odd functions.

In particular this is true for all $f\in L^1\cap L^2$ and by continuity it is true for all $f\in L^1$. (In fact an analogue statement is true for all tempered distributions.)

EDIT: Oh, I just saw that you asked for the other direction. Using the same argument you can show that $\mathcal{F}f$ odd $\implies f=\mathcal{F}^3(\mathcal{F}f)$ odd is true for all $f\in L^2$. This time I'm not quite sure if it is possible to extend this from $L^1\cap L^2$ to $L^1$, but maybe the result for $L^1\cap L^2$ is useful for you too.

Comment by Johannes Hahn

2013-05-14T15:26:23Z

@Vladimir: I had no intention to be dismissive of your post and I apologize if it sounded like that. I simply wanted to correct what I though was an erroneous statement. I misinterpreted what you were referred to when you wrote that "this" was proved by Burnside. The comments (one of them deleted by now) before that were referring to a finite subgroup of G and it is certainly false, that f.g. subgroups of G have to be finite. Only after passing to the image in $GL(V)$ this is true and that's what I wanted to point out.

Comment by Johannes Hahn

2013-05-14T12:15:11Z

This was exactly not proved by Burnside. The assertion that a finitely generated group with $x^n=1$ for all elements x is finite, is exactly Burnside's problem and is known to be false for large enough exponents. However: This might just work out if we do not work in $G$ itself, but in the image of $G$ in $GL(V)$, because linear groups are much better behaved than general infinite groups when it comes to these kinds of problems. I do not know off the top of my head whether or not Burnside's problem is true for linear groups or not.

Comment by Johannes Hahn

2013-05-04T14:10:56Z

I don't think this is a good example for what the OP had in mind. The proof is maybe not nice to write down, but it is conceptually very easy. The very reason it is "obvious" is just because the proof is immediately obvious. The reason the proof isn't nice is rooted in the fact that Turing machines are not nice because of their minimalistic definition. If one would choose to define a theoretical computation machine with abilities more akin to a real world computer, then writing down the proof would reduce to some easy programming.

Comment by Johannes Hahn

2013-04-30T23:51:41Z

Well they are certainly "legal" but that's not the point. MO isn't the place for homework, it's for research level questions. Try math.stackexchange.com as an alternative.

Comment by Johannes Hahn

2013-04-12T17:54:00Z

One is tempted to say: In the representation theory of the symmetric groups! ;-)

Comment by Johannes Hahn

2013-04-11T12:47:01Z

Well there is the obvious advice: Try another book! Also: This is not the kind of question that is appropriate for MathOverflow. Consider asking it on other sites like math.stackexchange.com.

Comment by Johannes Hahn

2013-04-06T15:01:09Z

Even more generally the subrgroup $Aut(G)\leq Sym(G)$ lies in the normalizer of the cayley-embedded $G\hookrightarrow Sym(G)$ and thereby on obtains an canonical embedding $G\rtimes Aut(G) \hookrightarrow Sym(G)$ for all groups G.

Comment by Johannes Hahn

2013-03-26T17:42:31Z

The crucial point in your question on which you failed completeley to elaborate is obviously the question what do you mean when you say "really" ? In particular: Which part(s) of the proof of the completeness theorem is (are) questionable in your view? Unless you clarify your question, it does not constitute a real question for MO ;-) (And a minor point: It's not "the" but rather "a" model that exists by the completeness theorem. There is no uniqueness claim made.)

Comment by Johannes Hahn

2013-03-24T18:47:00Z

The $F_4 \to E_6$ embedding can be obtained by a process of "folding" at least at the level of Weyl groups. I'm pretty certain that this should be possible for lie algebras as well. In essence $\mathfrak{f}_4$ should be the space of fixed points under the diagram automorphism of $\mathfrak{e}_6$.

Comment by Johannes Hahn

2013-03-24T00:41:40Z

There is a "Hausdorffication": Every topological space has a biggest hausdorff quotient, that is a quotion $X\to HX$ such that all continouos maps $X\to Y$ into hausdorff spaces factor as $X\to HX\to Y$.

Comment by Johannes Hahn

2013-03-09T18:45:21Z

"Homoemorphism" is the wrong word, there are no topologies involved here. If you change you definition of $\hat{g}$ to $\hat{g}(\omega) :=x\mapsto\omega(g^{-1}x)$ then everything works out and you get indeed a group action of $G$ on $\Omega$. Also: This is not the right place to ask this sort of question. MO is for research level questions. Try math.stackexchange.com for example.

Comment by Johannes Hahn

2013-03-07T18:16:21Z

@some1.new4u: Then perhaps math.stackexchange.com is the more appropriate site for you to ask your questions. MO is for research-level questions.

Comment by Johannes Hahn

2013-03-06T00:02:09Z

Fu... I just "proved" that again as an exercise a few days ago.

Comment by Johannes Hahn

2013-03-05T17:52:30Z

I know that the Hessian isn't exactly the differential of something, but shouldn't there be an analogue to closedness of forms for it? Then there would be topological obstructions for the implication "closed => exact" of course.

Comment by Johannes Hahn

2013-03-05T17:44:29Z

Am I the only one who wonders why the second cohomology is singled out? This somehow raises the question whether there are any natural actions on single cohomology groups that do not come from actions on the whole cohomology.