Algebraic Topology - Science topic

I guess operads are not so popular.

@peterkepp

I have started reading algebraic topology from three book

Topology by J Munkres

Algebraic Topology by A. Hatcher

Basic of Algebraic topology by A. SHASTRI

But I can confused from where did I find the research problem and the recent papers are too advance for me

A list of open problrems for hpmotopy type theory is presented in HISTORIC.

Stiefel Manifolds and Grassmann manifolds are two examples of Smooth/analytic Riemannian Manifolds, with its far reaching applications in data analysis, such as classification, clustering and object tracking etc.

Hi Clarence Lewis Protin

I recommend the website http://128.2.67.219/homotopytypetheory/show/open+problems

Regards,

Germán Benitez Monsalve

Clarence Lewis Protin I had a look on Wikipedia to find out about Thom-Mather stratified spaces:

It mentions its use in the study of singularities. My comment about trying to apply this approach to cosmology is based on the Spacetime Wave theory:

From this worldview, singularities do not exist in physics and the laws of physics apply everywhere and for all time. Also the idea that spacetime may have a sponge like or fine grained structure (quantum fluctuations in empty space) is ruled out by the adoption of the Einstein equations of GR as the fundamental equations of spacetime at all scales. This means that if the Mass Energy tensor is identically zero then spacetime curvature must be identically zero.

This the idea of quantum fluctuations in empty space from quantum theory has to give way.

Richard

Agreed with J. G. von Brzeski

Hamed Sadaghian Thank you, this is very useful for me.

see

1-A C n -MOVE FOR A KNOT AND THE COEFFICIENTS OF THE CONWAY POLYNOMIAL

OHYAMA, YOSHIYUKI, YAMADA, HARUMI

Journal:

Journal of Knot Theory and Its Ramifications

Year:

2008

2-Concordance invariance of coefficients of Conway's link polynomial

Tim D. Cochran

Journal:

Inventiones mathematicae

Year:

1985

see

congratulations

The difficulty depends on your view of the (co)homology groups. Generally speaking, explicit formulas involving (co)cycles are not of great use. The best approach is to go back to the fundamental property of the (co)homology functor, as is done in Serre's "Local Fields", chap.VII, which is in my opinion the best introduction to the subject.

Let us look e.g. at homology. Given a group G and its group algebra A = Z[G], an exact sequence of Z[G]-modules o --> A --> B --> C --> 0 gives rise to an exact sequence of co-invariants ... --> A_{G -->} B_G --> C_G --> 0, where H₀ (G,M) _:= (M)_G denotes the maximal quotient of M on which G acts as identity. But this sequence is not exact on the left, where it extends to an a priori infinite exact sequnce of homology groups ... --> H₁(G,A) --> H₁(G,B) --> H₁(G,C) --> H₀(G,A) --> ... The philosophy is that the homology functor is an automatic machinery which transforms a short exact sequence into an infinite exact sequence on the left. Think of the Taylor development of a function f at a point M, which you can write down without thinking, but the terms of the expansion give you valuable information on f itself in the neighbourhood of M.

1) Let us apply this to your specific problem. Denote by I_G the augmentation ideal of A , i.e. the ideal generated by all the elements 1-g when g runs through G. Obviously H₀(G, M) = M/I_GM . Consider the natural exact sequence 0 --> I_G --> A --> Z --> 0, where G acts by conjugation on G and as identity on Z, and the rightmost map consists in taking the sum of the coefficients of an element of the group algebra A. Taking homology gives ... --> H₁(G, A) --> H₁(G, Z) --> H₀(G, I_G) --> H₀(G,A) -->... But A is a free A-module, hence H₁(G, A) = 0, and the image of H₀(G, I_G) --> H₀(G,A) is null practically by definition, so that H₁(G, Z) = I_G/ I_G² , and one can check "by hand" that the map g --> 1-g induces an isomorphism H₁(G, Z) = G/G' (remember that G acts by conjugation on itself).

Consider now the exact sequence 0 --> Z --> Z --> Z/n --> 0, where the leftmost map is multiplication by n , hence is injective.Taking homology gives H₁(G, Z) --> H₁(G, Z) --> H₁(G, Z/n) --> Z --> Z . Applying our preliminary result H₁(G, Z) = G/G' , we get immediately that G/G'Gⁿ = H₁(G, Z/n) as desired.

2) Suppose now that G is a finite p-group, where p is a prime, and write F_p instead of Z/p. After 1), we know that G/G'G^p = H₁(G, F_p). Both sides are finite abelian groups killed by p, so they can be viewed (when written additively) as F_p finite - dimensional vector spaces. Let d(G) denote the minimal number of generators of G. Passing to the quotient, we get obviously d(G) >= d(G/G'G^p ), and this last quantity is = dim G/G'G^p by elementary linear algebra.

The reverse inequality holds true, but its proof is somewhat independent from the previous homological considerations. Write G*=G/G'G^p. A theorem of M. Hall in group theory implies that a homomorphism f : G₁ --> G₂ is surjective iff the induced map f* : G₁* --> G₂* is surjective. The equality d(G) = dim G/G'G^p is known as the Burnside basis theorem.

I think that you can use the article

FROM HOMOLOGICAL ALGEBRA TO GROUP COHOMOLOGY

Semester Project By

Maximilien Holmberg-Péroux

Responsible Professor

prof. Jacques Thévenaz

Supervisor

Rosalie Chevalley

Academic year : 2013-2014

Spring Semester

James F Peters, Artur Sergyeyev Thank you very much Professors for your king suggestions. However, I am looking for where I can get a free e-copy of the book.

The most common examples of vortical structures are found in classical fluids; at the laboratory scale, we can mention

In the rotating stratified fluids in geophysical systems, we have a multitude of vortices at different scales; they include

The main characteristic of all these structures is that they do not keep their identity for a very long time, the main exception being perhaps the red spot of Jupiter which has preserved its integrity over the last few hundred years.

The completely new reality of vortex systems is in quantum fluids where they can last indefinitely.

On a larger scale, in space, practically all forms of matter organization come in the form of swirls with the most spectacular examples provided by rotating galaxies. It is fairly easy to conclude that the vortex movement is ubiquitous in all physical systems, from the smallest to the largest. There are even some attempts to connect the scales proposed in the framework of a Cantorian Superfluid Vortex hypothesis; these attempts are very similar, in many respects, to Lord Kelvin's old hypothesis explaining atoms as vortex structures.

For an in-depth discussion of all these fascinating topics, please see a very interesting article by Frank Wilczek entitled: "Beautiful Losers: Kelvin's Vortex Atoms". The text is available at:

The existence or non-existence or transitions between different kinds of particles may have small scale geometrical implications. This has nothing to do with a possible existence of gravitational impact of a curved large scale geometry.

@Romeo P.G

For sure this is true. There are many examples from algebraic geometry. I often work with kinds of locally ringed spaces - think of just generalising the sheaf of smooth functions on a manifold -  and these are not `set theoretical objects' in the sense they are not simply a set of points (together with a topology or something like that). 

But of course, as ringed spaces morphisms can be defined as so can their category. The point is that the objects are not sets, though in the cases I deal with, the Hom sets are genuine sets.

This is an important question with many possible answers.

A good overview of computation for persistent homology is given in

An approach to constructing Cech complexes is given in Section 2.2.1, starting on page 5 in

See pgs 102-104 of Hatcher's "Algebraic Topology" for a precise definition of Delta complex. It roughly says that the interior of any simplex is mapped homeomorphically to its image in the space, though interiors of different faces may be identified.

This is a great question.

For an answer to this interesting question, see

arXiv:1611.07441v1

at

For periodic variants of the square peg problem, see Section 4, starting on page 12.

For the algebraic square peg problem in a 2014 M.Sc. thesis, see

arXiv:1403.5979v1

also

at 

Look at this doc it may be helpful for your topic. Good luck.

It seems to me "germs of homeomorphisms at 0"

A little comment. In the definition of local homeomorphism  f:U--->R^n there is no need to say that f is open. Openness follows from continuity and injectivity, according to  the Brower invariance of domain principle. See the exposition written by Terry

Tao in his blog:

A lot of good answers, I'll try to explain intuitively why this is the case, but please take this with a grain of salt, as I'll be arguing heuristically.

At a very fundamental intuitive level, a homeomorphism h: X -> Y, is the requirement that there is a correspondence between open sets in the topological space Y correspond precisely to open sets in the topological space X (and vice versa).

If you think back to the definition of a topological space (from which all of the topological properties arise), it is entirely dependent on which open sets exist in your given space, and what properties arise as a consequence, such as compactness or connectedness. Consider the two extreme cases - If you were to look at |R under the discrete topology (every possible subset is declared an open set) this is non-compact, and separable while |R under the trivial topology (only |R and the empty set are considered open sets) this is compact and connected.

In your specific question, the metric topology can be induced on space X... and hence you'd like to know if a homotopy is sufficient to induce metrizability on space Y. However, we would need some correspondence between the induced open sets of Y and the open sets of X to make sure that we don't somehow "reduce/increase the number" or somehow fundamentally disrupt the behaviour of open sets (very loosely speaking, since we're dealing with issues of countability here).

This condition necessarily means we have a homeomorphism from X to Y...and since we can't guarantee this with homotopy alone, we cannot guarantee X and Y have the same topological properties.

And just to add an explicit example, as Vladimir said above, let X be D^2, the unit 2-D Disk, and {x} be the origin of \R^2. There exists a homotopy that maps X to {x} in a continuous manner (imagine X 'shrinking' forever until a single point remains in the limiting case). But clearly, no homeomorphism exists as these are very different topological spaces. (Such spaces X are said to be contractible)

But in the typical reflection the kinetic energy goes to zero at the wall while the potential energy is at a maximum so there is a force that explains the reflection,  If a wave travels down a funnel it does not reflect.  So I say these imaginary traveling ways are nonsense.  Where is the equation of motion if the wave is standing?  Where is the stroboscopic image of these two waves that some how reflect to make one.  There is a curve lifting function between the string at rest and the string in wave form that is point-for-point so the only force vectors are orthogonal to the string axis.

Statements about the string without quantifiers are provable.  Can you prove there is a mode lower than the fundamental with wavelength 2L? Of course not.

Dear  Terence B Allen, 

I need to understand better your question.

At the moment I note only if we apply the algebraic rule for complex multiplication we have

f(z)= z^2 =(x + iy)(x + iy)= x^2 - y^2 +2ixy. This mapping maps the circle S^1 onto itself. But if z travels along the circle S^1 one times then the image

f(z)= z^2 goes two times around S^1. We can also write f(e^{i\theta})= e^{2i\theta}, where \theta is the measure of the angle.

Boundedness is a topic of bornology. You can introduce many bornologies on your spaces, so you will get many notions of a bounded set.

Dear Terence, my limited knowledge doesn't permit me to speak about those "music spaces". I am just surprised by a statement you made: """Clearly the torus is isomorphic to a closed, bound rectangle in a plane""". Not at all. Consider a rectangle ABCD. To get a torus, one has to identify AB with DC, and then identify the two circles such that A  is identified with B. So you get ABCD / rel, where rel is an equivalence relation, and the result of this factorisation of a topological space is not at all an isomorphic (homeomorphic) topological space. For example, let M be the middle of AB translated with epsion to the right side and N be the middle of DC translated with epsilon to the left side. The rectangle \setminus MN is no more a connected topological space, while the torus \setminus MN is still connected. Of course, there are thousands of other reasons why they are not isomorphic. Similarly, rectangle, torus, sphere and projective space are pairwise not isomorphic (not homeomorphic). 

The answer is No. It is a standard fact that the complete graph K_5 on 5 vertices (i.e.a_0, ...,a_4 with all pairs of vertices joined by edges) is non-planar, i.e. it does not admit an embedding in the plane. This is also true for the complete bi-partite (3,3) graph.  It is a non-trivial theorem of Kuratowski that any non-planar graph contains one of these two as a subgraph. See the wikipedia article https://en.wikipedia.org/wiki/Kuratowski%27s_theorem and references therein.

I am a professional musicologist and music theorist, and I must confess that what you people write is Chinese to me. Music can be notated in staff notation, or in various kinds of tablature, or in various types of alphanumeric notations. Passing from the one to the other (which is called "transcription") can now be automated, especially with the help of MusicXML and MEI (Music Encoding Initiative).

BUT this all is valid only provided that the music can be notated at all. And the transcriptions exclusively concern the notated aspects of the music. Notation is utterly unable to notate everything. What it notates is called "notes", which represent a combination of pitch and duration. But, because notations (like music itself) are semiotic systems, the notated pitches and durations are but abstract categories escaping any attempt at quantification. Notes are distinctive elements of a semiotic system. And, in addition, not every music can be notated, which probably means that not every music is semiotic.

The so called pitch classes of musical set theory are abstractions. If they are considered classes properly speaking (which I doubt they are), then the members of these classes must themselves be considered abstractions, and so also are the intervals between them. And none of this can be evaluated in terms of "true/false". Musical set theory is not a set theory in the sense of Zermelo-Frankel.

Take a look at the GUDHI library: http://gudhi.gforge.inria.fr/doc/1.2.0/

based on Simplex Tree: An Efficient Data Structure

for General Simplicial Complexes (http://www-sop.inria.fr/members/Clement.Maria/articles/Boissonnat,%20Maria%20-%20The%20Simplex%20Tree.pdf)

Instead of an answer I have an example. Let A be nonmeasurable in X, f:X->X, f(X)=Z f(x)=x, for x in Z, f(x)=z, z in Z. Then f^{-1}(Z) =X. The preimage of nonmeasurable set is measurable.

Hello.

Handbook of Algebraic Topology edited by I.M. James, pages 561-566 may have what you want.

See also the notes by Galatius at http://math.stanford.edu/~galatius/notes3.pdf for recent work.

Dear Ion C.Baianu

For me a groupoid is a small category in which all morphisms are invertible. Would you please provide the explicit definition of omega groupoid?

Maybe you would like to see geometrically how the n-th homology of  S^n can be killed when projected to RP^n? Take a triangulation of the n-sphere which is invariant under the antipodal map  -I , e.g. the generalized octahedron (which exists in every dimension). The projection is a triangulation of the projective space RP^n, and each of its simplices is hit twice by the projection map S^n \to RP^n. However, when n is even, the antipodal map -I on R^{n+1} reverses orientation (determinant -1) but preserves the exterior normal vector field of S^n, thus it reverses the orientation on S^n. Hence each simplex contributes with both signs and you get extinction, while in odd dimensions you obtain a factor 2 (the mapping degree of the projection map).

Best regards

Jost

Since n>1, we have that π_1(S^n)=0. Thus, by the lifting criterion for covering spaces, the map f : S^n → M lifts to a map f~ : S^n --> M~, where M~ is the universal cover of M. 

Now, if π_1(M) were to be infinite, the universal cover of M would be non-compact, in which case H_n(M~)=0.  But this would force the degree of f to be zero, a contradiction.

Dear Nounahon,

                               I would suggest you take a look at my papers on Haag Theorem and Politcial Environment and Stock Markets papers on my RG page. They might be helpful. Although I do not know what exactly you mean by topological twist is it conformal flows or heterotic flows that you are looking for? SKM QC FEPS(D)

This is a good question that taps into cutting edge research on applications of the usual notion of homotopy considered in the context of sources of digital information sources such as digital images.

A good place to start in answering this question is in

Sang-Eon Han, KD-(k0, k1)-HOMOTOPY EQUIVALENCE AND ITS APPLICATIONS, J. Korean Math. Soc. 47 (2010), No. 5, pp. 1031–1054:

Digital homotopy is defined in the context of the Khalimsky digital topological category defined by

1.  class of objects

2.  KD-(k0, k1)-continuous maps as morphisms.

A map f : X → Y is KD-(k0, k1)-continuous at a point x0 ∈ X, provided

(1) f is continuous at the point x0; and

(2) for any Nk1 (f(x0), ε) ⊂ Y , there is a neighbourhood Nk0(x0, δ) ⊂ X such that f(Nk0(x0, δ)) ⊂ Nk1 (f(x0), ε), where ε, δ ∈ N.

The map f : X → Y is KD-(k0, k1)-continuous on X into Y, provided it is continuous at every point x in X.   From this, we obtain the notion of a digital homotopy.  Then we say that F is a digital KD-(k0, k1)-homotopy between f and g, and f and g are KD-(k0, k1)-homotopic in Y.

Dear Eliza,

Regarding your question about  Horst's mistake I believe that it was just a typo. The reason I believe this is that  I have proved this result and I have communicated it to him. See the the reference ker 200? in the references in his book. Unfortunately the status of the paper is still unknown. Recently Paul Howard ask for copy of this paper and I have prepapred a pdf file which I will forward to you via e-mail. I think that Horst told me once that some guy from Brazil also notice this typo. However, I do not know whether he has publish anything.

What is meant by the Euler-Poincare principle in the text is that in a complex of finite dimensional vector spaces the alternating sum of the dimensions of the spaces coincides with the alternating sum of the dimensions of the homology groups. (The standard application of this is that the Euler characteristic of a finite CW-complex coincides with the alternating sum of the number of cells in each dimension.

If you complex is not just a complex of vector spaces but a complex of representations of a semi-simple Lie algebra $\mathfrak g$ with $\mathfrak g$-equivariant differentials, then you can look at weight spaces in the representations. Since all differentials map weight spaces to weight spaces, you can apply the above argument for each fixed weight (provided that all weight spaces are finite dimensional). So for each weight, the alternating sum of the dimensions of the weight spaces in each homology group coincides with the alternating sum of the dimensions of the weight spaces in each chain group. Since the character is just the formal sum of the dimensions of the weight spaces, it follows that the alternating sum of the characters of the homology groups coincides with the alternating sum of the characters of the chain groups.

Eric is right. I am a finite element person and don't know much about geometry which is tooooo hard for me.  What we are doing is to divide a 3D convex polyhedron domain to many small convex polyhedrons shared faces. Then define piecewise polynomials associated with these partitions (or meshes) to approximate solutions of PDE.

Let me make it straight that for my situation, the formula is

n_0 − n_1 + n_2 − n_3 = −1.  

with  

n_0=total  number of interior vertices,

n_1= total number interior edges,

n_2=total number interior faces

n_3=total number polyhedrons (not just interior). 

Is the above statement right?

Many thanks to Stam and Eric.

I would like to also recommend chapter 2 ("Global Fields") of the classical book 3Algebraic Number Theory", Cassels & Fhröhlich ed., Acad. Press 1969, which gives a compact and complete account of the subject. Concerning your specific question, see especially §7, p.50

 Dear Yurii, I am not sure I understood your question since the number of intersection points between any 2 closed curves in R^2 (counted with multiplicities) is even. Thus your condition is automatically satisfied. 

Yours B.Shapiro

But your topological  consideration  on G is  a motivation to ask the following question:

Question: Assume that G is  a topological group. Is there a CW complex X wich fundamental group is isomorphic to G and  the  standard  action of G on $\tilde{X}$, the universal covering space of X,  is jointly continuous in two variables,  that is the action as a map \tilde{X} \times G to  \tilde{X}  would be continuous.

Just a remark. (I write it for singular cohomology, but it should be similar for simplicial homology).

On the cochain level we have:  0-> C^*(X,A)->C^*(X)->C^*(A)->0, (where C^*(X,A)=Hom(Delta_*(X)/Delta_*(A),\Z) etc.). A relative cocycle (from C^*(X,A)) is represented in C^*(X) as an absolute cocycle that vanishes on A. 

So what is 'absolute cocycles modulo relative boundaries'?

A special case gives an idea of what the result should be: If A is a component of  X, every absolute cocycle can be uniquely written as the sum of (the image of) of a relative cocycle and cocycle on A; taken the quotient with respect to the relative co-boundaries, we get an element of H^*(X,A) + Z^*(A),  i.e. the sum of a relative cohomology group and a cocycle group.

In general (if A is not a component of X) I would expect something like H^*(X,A) + some subgroup of Z^*(A) (didn't think about it, though).

Thus the quotient you are looking for is not a true 'cohomology'; in the setting of simplicial cohomology it would depend on the cell decomposition.

The Lie algebra of smooth vector fields on a manifold has no finite dimensional ideals.

Proof. Assume that g is a finite dimensional linear subspace in the Lie algebra of vector fields over the manifold M. Then M has a finite number of points p_1,\dots,p_k such that for any X\in g, X\neq 0, at least one of the vectors X(p_1),\dots,X(p_k) is not zero. (This follows from a compactness argument on the unit sphere of g with respect to an arbitrary positive definite inner product.) Choose a vector field X\in g, X\neq 0 and a point q, such that q is different form  p_1,...,p_k and  X(q)\neq 0. Then there is a vector field Y on M such that [Y,X](q)\neq 0. Choose a smooth function h such that h vanishes on an open subset containing the points p_1,...,p_k and h is constant 1 on an open neighborhood  of q. Put Z=hY. Then the Lie bracket [Z,X] is not zero, since

[Z,X](q)=[Y,X](q) \neq 0, on the other hand, [Z,X](p_i)=0 for i=1,...,k, thus, [Z,X] cannot be in g by the choice of the points p_1,...,p_k. This means that g cannot be an ideal. 

The sensitive dependence on initial conditions.

Definition. A metric S-system (X,d) is sensitive if it satisfies the following condition: there exists a (sensitivity constant) c > 0 such that for all x∈X and all δ>0 there are some y∈B_δ(x) and s∈S with d(sx,sy) > c. We say that (S,X) is non-sensitive otherwise.

In my opinion, if the product system of f and g be mean sensitive then f or g is need not sensitive.

Please see the attachment.

It the product system of f and g are strongly sensitive then both the systems f and g will be so. However, I doubt the result in case of snydetically sensitive. It would be interesting to see an example here.

Take any automorphism of a compact topological space X that has a non trivial action on homology, cohomology or any other homotopy invariant algebraic topological invariant.  Then the free group generated by the automorphism acts on X and so it acts on C^0(X). But clearly the Z action cannot be extended to an R action, because that R action must come from an R action on X [1]. Such an R action would be homotopically trivial contradicting that the generator of the Z action gives a non trivial algebraic topological invariant. 

Example: the action of the mappingclass group on a two dimensional oriented surface Sigma  which give non trivial elements of  Sp(2g,Z). In particular the so called Dehn twists, give examples with non trivial actions on H^1(Sigma). 

Now obviously this is not a non commutative example, but if I understand correctly the equivariant case (a topological space with a toplogical group action) gives rise to non commutative C^* actions that are related to the space with a a group action in much the same way as above, i.e look for a space with an equivariant automorphism that gives rise to a non trivial element in equivariant (co) homology or other equivariant invariants. 

[1]  The Gelfand Naimark theorem is not just that the statement about maximal ideals of commutative C^* algebras correspond to points, but also that continuous *-homomorphisms correspond to continuous maps. Its a functorial equivalence of categories!

[2]  the Z- rank of H^1(Sigma) is 2g which comes with non degenerate symplectic form defined by cup product and evaluation on the fundamental class

[Sigma] \in H_2(Sigma).

M^eq is not affine; it is an expansion of the vocabulary to guarantee that quotients are present. For definition of terminology see http://en.wikipedia.org/wiki/Imaginary_element

The examples there of `elimination of imaginary' are dull; more interesting is algebraically closed fields.

The compactness theorem is sometimes used to reduce infinite coverings to finite.  See for example Poizat's stable groups for the interaction between definable converings and algebraic geometry.

In [Hilton: A new look at means of topological spaces (http://www.emis.de/journals/HOA/IJMMS/Volume20_4/584598.pdf)], the dual notion of co-mean would provide such a morphism in the category of commutative R-algebras (the co-product being the tensor product in this category.)  I don't know whether this carries over to commutative  Banach algebras (is the coproduct in this category  the completion of the tensor product?)

Yes, and you can smooth out the map to count preimages of regular values, and connectivity needs to be assumed. Not clear, though, whether S can have aspherical type over Z₂ or Z, as this may take more than algebraic topology even in 3D.

Pozdrawiam, Witold.  

Dear Aderemi,

I believe a positive answer to your question can be derived for Azumaya algebras using the results of Dwyer and Friedlander in their paper:  Étale K-theory of Azumaya algebras.  Proceedings of the Luminy conference on algebraic K-theory (Luminy, 1983).  J. Pure Appl. Algebra 34 (1984), no. 2-3, 179-191.  

@ Muhammad Sirajo; If |z_j|<1 and |w|=1, it is not necessary that |w-z_j|<1. For example if w=1 and z_j=-1/2 then |w-z_j|=3/2>1.

Dear sir,

Please refer to this link : http://en.wikipedia.org/wiki/Binary_tree#Combinatorics

The number you look for is related to Catalan number. You should tag your question with : "Combinatorics" and/or "Combinatorics on words". You will find more people that will help you to get the right algorithm.

Best wishes.

You're welcome. Take a look at the article from http://www.iucr.org/__data/assets/pdf_file/0006/13929/final_23.pdf

Besides having nice visualizations, he seems to put his graphical routines at public disposal. His author, de Baer also wrote a book with McHenry: Introduction to Crystallography, Diffraction and Symmetry (see https://books.google.ch/books?id=NMUgAwAAQBAJ&pg=PA223&lpg=PA223&dq=visualization+of+magnetic+groups&source=bl&ots=VQ3MagtpF8&sig=oEbp78MRrjxPfdt2D1WUpDY5cwM&hl=en&sa=X&ei=ucjTVIjrFI3SaJjLgKgB&ved=0CEwQ6AEwBw#v=onepage&q=visualization%20of%20magnetic%20groups&f=false).  You might also want to look at http://www.cryst.ehu.es/

Good luck!

You cn also look the definitions of Balanced and regular categories. Infact GRP is a regular category and so bimorphisms are isomorphisms.

thanks.

The analogue you're looking for is the notion of tangent groupoid. You can look it up in the book of A. Connes. Also in several other places -- it is central in Noncommutative Geometry. As for stratified pseudo-manifolds, it is used there as well (see work of Claire Debord and Jean-Marie Lescure).

we can define a map Fg:G--->G by Fg(x)=gxg^-1.

then it will be a homomorphism and also continuous map .

In your previous question you asked if every neighborhood of the neutral element is absorbing. You have received a number of answers with various counterexamples. Every such a counterexample works also as a counterexample to your new question: if g is not absorbed by any Vⁿ then gU is not absorbed neither.

Moreover, for your new question the answer is negative even for locally convex Hausdorff topological vector spaces  (considered as additive groups). Namely, let X be the space. Take U = X, then g+U = X, and surely there is a convex neighborhood of 0 V such that for every n  the set nV is not equal to the whole X.

I would just like add a few remarks to the first part of M. Amiri's answer: 

Perhaps one should explain which quantities in classical physics can be represented by 1-forms at all, because at first sight they should look like vector fields. Dual vectors and vectors are related by a pairing. Given a physical quantity that is modelled as a vector field, and given a pairing with another quantity, the latter should be modelled as a 1-form.  E.g. velocites are modelled as vector fields. There is a pairing between velocities and forces, yielding the power needed (or obtained)  when you move with velocity v in the force F. If you prefer not to express this pointwise you might say that there is a pairing between differentiable paths and force fields F, yielding the energy (used or obtained) on the path in F.     With respect to this energy/power-pairing force fields are a typical example of physical quantities that are represented by 1-forms in a natural way. Closed 1-forms are 'locally conservative force fields',  exact 1-forms are 'conservative force fields'. 

That's the nice part of the story, but the geometrical meaning of quantities is often obliterated in the physics idiom:

Of course, given a 1-form \alpha and a non-degenerate bilinear form (e.g. a (semi-) Riemannian metric g(.,.)) one can define a vector field w that represents \alpha via g(w,v)=\alpha(v) for all v.  

Phycicists would normally say that under coordinate changes the coordinate vectors of vectors  behave 'contravariantly', say via x'=Ax, and those of dual vectors 'covariantly' via l'=(A^T)^{-1} l.  The possibility to represent a dual vector (a 1-form) by a vector (a vector field), given such a nondegenerate bilinear form g, goes like this in the physics language:   The same quantity has contravariant and covariant coordinates.

In Euclidean 3-space 'the coordinates of a force' are the coordinates of a 1-form \alpha with respect to the dual basis (e^*_1,...,e^*_3)  to some reference orthonormal basis (e_1,...,e_3) .  The exterior derivative d\alpha is a two form which has again three coordinates with respect to  the basis (e^*_2 \wedge e^*_3, -e^*_1\wedge e^*_3, e^*_1\wedge e^*_2).   With respect to these choices of basis exterior differentiation is the 'curl'.

---

Dillon Scofield gives an important example of a closed  (exact) 2-form in physics, not a 1-form. The electromagnetic field F is a 2-form on 4-d-space.

Hi Edson. You are right, in 1) and 2) instead of

$ r = ... = topdeg(g \circ id|_{L'}) \leq topdeg(g|_{L'}) = algdeg(g) = k$

there must be

$ r = ... = topdeg(g \circ id|_{L'}) = topdeg(g|_{L'}) \leq algdeg(g) = k$

I corrected the answer.

On the other hand $r$ can be not equal to $k$, that is it may happens that

$r = topdeg(g|_{L'}) < algdeg(g) = k$.

It seems that in this case we always have that $r=0$, but I can not produce at least just now an example when $0<r<k$.

For $r=0$ an example can be constructed as follows.

As in 2a) suppose that $L$ is contained in a subspace $\{ z_1 = 0 \}$.

Take $g(z_1,z_2,...,z_n) = z_1^2$.

Then $g|_{L'} = 0$, and so $0 = topdeg(g|_{L'}) < algdeg(g) = 2$.

As a first iteration, I would try this http://dml.cz/bitstream/handle/10338.dmlcz/107982/ArchMathRetro_042-2006-1_7.pdf ... It took me to this one http://journals.impan.pl/cgi-bin/doi?ap82-3-6 where is a reference [6]. This might be what you are looking for. However, the fiber preserving only.  When more general, this should be mentioned in [6] (Kolar, Mikulski DGA 1999. I guess the biggest group working in this area contains Doupovec, Kurka, Miukulski and Kolar (they should be active in this field...as far as I know). Of course, it depends in waht extent functors you want to classify ... and how fine and explicit is the classification-- often thse are done using the algebraic (Weil algebra)  or vector space data.

It does not hold when you drop the assumption of convexity. The first planar counter-example was found by Klee:

V. L. Klee, Some topological properties of convex sets. Trans. Amer. Math. Soc. 78, (1955). 30–45.

see also

V. L. Klee, An example related to the fixed-point property. Nieuw Arch. Wisk. (3) 8 1960 81–82.

Connell found other examples:

E. H. Connell, Properties of fixed point spaces. Proc. Amer. Math. Soc. 10 1959 974–979.

In particular, he showed that FPP is not closed under finite products or taking closures.

However, for convex sets in Rⁿ your question has an affirmative answer. See p. 35 in Klee's paper from TAMS and p. 958 in Connell's paper from PAMS.

What about motivating intuitions? Topology was developed basically to deal with intuitions about "space," "connectivity, "continuity," notions of "near" and "far," etc. Algebra came about in order to deal with notions of "finitary manipulation," especially in connection with equalities. The historical developments happened quite organically, without any particular "guiding hand," so the boundaries are fuzzy indeed...

@ Ulrich Mutze: Infinitesimal changes of state would define vector tangent to the state space. Alternatively one might define these such a vector as a local derivation on the set of state functions. Work is pointwise a linear form on the tangent space, i.e. the space of 'state change vectors'. For this 1-form (or covector) to be defined one would have to consider sufficiently many state changes to span the tangent space. In principle one can change all quantities, that can be considered as the coordinates of the state manifold. If we have extensive quantities V, N, S as state variables, only changes in V would contribute to mechanical work. The 1-covector pdV is zero when evaluated along submanifolds defined by chaninge S and N only. I think that even the practically minded physicist somehow notices that pdV says 'how much work is done along a process' and so implicitly acknowledges work to be defined by a 1-form / a covector field.

I don't know whether there is a natural Riemannian structure in thermodynamics which would allow to give an interpretation of this 1-form as a vector field.

A Clifford algebra is defined for a quadratic vector space - so it's a purely algebraic object not depending on a manifold. But if you have a spin manifold M, then, depending on the topological structure, it can admit several spin structures (they are classified by H^1(M;Z_2)), and each spin structure yields a spinor bundle, i.e., a bundle over M that looks *locally* as the spin representation constructed via the associated Clifford algebra, but globally not. For details, look at Th. Friedrich's book

"Dirac operators in Riemannian geometry" (AMS, GTM vol. 25).

To Behrad Mahboobi: Just to reinforce my response to your question that we cannot say that an acyclic space is necessarily contractible, I thought I should let you know there is a result which says that a space X is contractible if and only if it is acyclic and its fundamental group is trivial. So an acyclic space is not contractible unless it has a trivial fundamental group.

You can also read this notes on Hodge theory: Foundation of differentiable manifolds and Lie group ( F. Warner) and form differentiali et loro integrali (G. de Rham).

I think the prove can be done along the following lines:

There is an injective immersion (i.e. embedding) $\phi: \partial M\to R^{2n+1}$ by the (weak form of) Whitney embedding theorem (2(n-1)<2n+1). Extend this map to an embedding $\Phi:\partial M\times [-\epsilon,\epsilon] \to R^{2n+2}$ by $\Phi(x,t)=\phi(x)+t e_{2n+2}$.

Identify (by means of a Riemannian metric) $\partial M\times [-\epsilon,\epsilon]$ with a collar neighbourhood $N$ of $\partial M$ in $2M$. Thus we obtain an injective immersion $N\to R^{2n+2}$ which we extend to a smooth map $\Psi: 2M \to R^{2n+2}$. The extension can be chosen in $E_+$ for $M_1\setminus N$ and in $E_-$ for $M_2\setminus N$, where $M_1$ and $M_2$ denote the two copies of $M$ in $2M$. $\Psi(2M) $ intersects $\partial E_+$ in $\Psi(\partial M)$.

By the weak Whitney embedding theorem there is an injective immersion (i.e. embedding) arbitrarily close to $\Psi$ which is identical to $\Psi$ on $N$ (as 2n<2n+1$. This map is an embedding as required.

If there is no error in this argument it should work for $2M\to R^{2n+1}$. I don't know whether the claim would be correct for $2M\to R^{2n}$, but I suppose the Whitney trick would work in each of the halves.