User marcos cossarini - MathOverflow

Nontrivial copies of SO(r) in SO(n)

2013-01-25T18:54:30Z

If $G=SO(n)=SO(\mathbb R^n)$ and $r\leq n$, it is easy to find a closed subgroup $H\leq G$ that is isomorphic to $SO(r)$, just let $S\subseteq\mathbb R^n$ be an $r$-dimensional subspace and let $H=\{g\in G:(\forall x\in S^\bot)g(x)=x\}$ . These are the trivial examples.

If $r=2$ and $m\geq 4$ we can find nontrivial embeddings $SO(r)\to SO(n)$, for example:

$\begin{pmatrix} \cos t&-\sin t\\ \sin t&\cos t\end{pmatrix} \mapsto \begin{pmatrix} \cos(2t)&-\sin (2t)&0&0\\ \sin(2t)&\cos(2t)&0&0\\ 0&0&\cos(3t)&-\sin(3t)\\ 0&0&\sin(3t)&\cos(3t)\end{pmatrix}$

Are there examples for $r=3$ (or more)?

Answer by Marcos Cossarini for Why is a topology made up of 'open' sets?

2010-03-27T17:58:52Z

In this answer I will combine ideas of sigfpe's answer, sigfpe's blog, the book by Vickers, Kevin's questions and Neel's answers adding nothing really new until the last four paragraphs, in which I'll attempt to settle things about the open vs. closed ruler affair.

DISCLAIMER: I see that some of us are answering a question that is complementary of the original, since we are trying to motivate the structure of a topology, instead of adressing the question of which of the many equivalent ways to define a topology should be used, which is what the question literally asks for. In the topology course that I attended, it was given to us in the first class as an exercise to prove that a topology can be defined by its open sets, its neighbourhoods, its closure operator or its interior operator. We later saw that it can also be stated in terms of convergence of nets. Having made clear these equivalence of languages, its okay that anyone chooses for each exposition the language that seems more convinient without further discussion. However, I will mantain my non-answer since many readers have found the non-question interesting.

Imagine there's a set X of things that have certain properties. For each subset of S there is the property of belonging to S, and in fact each property is the property of belonging to an adequate S. Also, there are ways to prove that things have properties.

Let T be the family of properties with the following trait: whenever a thing has the property, you can prove it. Let's call this properties affirmative (following Vickers).

For example, if you are a merchant, your products may have many properties but you only want to advertise exactly those properties that you can show. Or if you are a physicist, you may want to talk about properties that you can make evident by experiment. Or if you predicate mathematical properties about abstract objects, you may want to talk about things that you can prove.

It is clear that if an arbitrary family of properties is affirmative, the property of having at least one of the properties (think about the disjunction of the properties, or the union of the sets that satisfy them) is affirmative: if a thing has at least one of the properties, you can prove that it has at least one of the properties by proving that property that it has.

It is also clear that if there is a finite family of affirmative properties, the property of having all of them is affirmative. If a thing has all the properties, you produce proofs for each, one after the other (assuming that a finite concatenation of proofs is a proof).

For example, if we sell batteries, the property P(x)="x is rechargeable" can be proved by putting x in a charger until it is recharged, but the property Q(x)="x is ever-lasting" can't be proved. It's easy to see that the negation of an affirmative property is not necessarily an affirmative property.

Let's say that the open sets are the sets whose characteristic property is affirmative. We see that the family T of open sets satisfies the axioms of a topology on X. Let's confuse each property with the set of things that satisfy it (and open with affirmative, union with disjunction, etc.).

Interior, neighbourhood and closure: If a property P is not affirmative, we can derive an affirmative property in a canonical way: let Q(x)="x certainly satisfies P". That is, a thing will have the property Q if it can be proved that it has the property P. It is clear that Q is affirmative and implies P. Also, Q is the union of the open sets contained in P. Then, it is the interior of P, which is the set of points for which Q is a neighbourhood. A neighbourhood of a point x is a set such that it can be proved that x belongs to it. The closure of P is the set of things that can't be proved not to satisfy P.

Axioms of separation: If T is not T0, there are x, y that can't be distinguished by proofs and if it is not T1, there are x, y such that x can't be distinguished from y (we can think that they are apparently identical batteries, but x is built in such a way that it will never overheat. So if it overheats, then it's y, but if it doesn't, you can't tell).

Base of a topology: Consider a family of experiments performable over a set X of objects. For each experiment E we know a set S of objects of X over which it yields a positive result (nothing is assumed about the outcome over objects that do not belong to S). If you consider the properties that can be proved by a finite sequence of experiments, the sets S are affirmative and the topology generated by them is the family of all the affirmative properties.

Compactness: I don't know how to interpret it, but I think that some people know, and it would be nice if they posted it. (Searchable spaces?)

Measurements: A measurement in a set X is an experiment that can be performed on each element of X returning a result from a finite set of possible ones. It may be a function or not (it is not a function if there is at least one element for which the result is variable). The experiment is rendered useful if we know for each possible result r a set T_r of elements for which the experiment certainly renders r and/or a set F_r for which it certainly doesn't, so let's add this information to the definition of measurement. An example is the measurement of a length with a ruler. If the length corresponds exactly with a mark on the ruler, the experimenter will see it and inform it. If the length fits almost exactly, the experimenter may think that it fits a mark or may see that it doesn't. If the length clearly doesn't fit any mark (because he can see that it lies between two marks, or because the length is out of range), he will inform it. It is sufficient to study measurements that have only a positive outcome and a negative outcome, a set T for which the outcome is certainly positive and a set F for which the outcome is certainly negative.

Imprecise measurements on a metric space: If X is a metric space, we say that a measurement in X is imprecise if there isn't a sequence x_n contained in F that converges to a point x contained in T. Suppose that there is a set of imprecise measurements available to be performed on the metric space. Suppose that, at least, for each x in X we have experiments that reveal its identity with arbitrary precision, that is, for each e>0 there is an experiment that, when applied to a point y, yields positive if y=x and doesn't yield positive if d(y,x)>e. Combining these experiments we are allowed to prove things. What are the affirmative sets generated by this method of proof? Let S be a subset of X. If x is in the (metric) interior of S, then there is a ball of some radius e>0 centered at x and contained in S. It is easy to find an experiment that proves that x belongs to S. If x is in S but not in the interior (i.e, it is in the boundary), we don't have a procedure to prove that x is in S, since it would involve precise measurement. Therefore, the affirmative sets are those that coincide with its metric interior. So, the imprecise measurements of arbitrary precision induce the metric topology.

Experimental sciences: In an experimental science, you make a model that consists of a set of things that could conceivably happen, and then make a theory that states that the things that actually happen are the ones that have certain properties. Not all statements of this kind are completely meaningful, but only the refutative ones, that is, those that can be proved wrong if they are wrong. A statements is refutative iff its negation is affirmative. By applying the closure operator to a non refutative statement we obtain a statement that retains the same meaning of the original, and doesn't make any unmeaningful claim.

An example from classical physics: Assume that the space-time W is the product of Euclidean space and an affine real line (time). It can be given the structure of a four-dimensional real normed space. Newton's first law of motion states that all the events of the trajectory of a free particle are collinear in space-time. To prove it false, we must find a free particle that incides in three non-collinear events. This is an open condition predicated over the space W^3 of 3-uples of events, since a small perturbation of a counterexample is also a counterexample. Assuming that imprecise measurements of arbitrary precision can be made, it is an affirmative property. I think that classical physicists, by assuming that these kind of measurements can be done, give exact laws like Newton's an affirmative set of situations in which the law is proved false. I also suspect (but this has more philosophical/physical than mathematical sense) that the mathematical properties of space-time (i.e. that it is a normed space over an Archimedean field) are deduced from the kind of experiments that can be done on it, so there could be a vicious circle in this explanation.

Answer by Marcos Cossarini for real symmetric matrix has real eigenvalues - elementary proof

2013-01-12T20:42:45Z

Let me give it a try. This one only uses the existence of a maximum in a compact set, and the Cauchy-Schwarz inequality.

Let $T$ be a selfadjoint operator in a finite dimensional inner product space.

Claim: $T$ has an eigenvalue $\pm\|T\|$.

Proof: Let $v$ in the unit sphere be such that $\|Tv\|$ attains its maximum value $M=\|T\|$. Let $w$ also in the unit sphere be such that $Mw=Tv$ (which is like saying that $w=\frac{Tv}M$, except in the trivial case $T=0$).

This implies that $\langle w,Tv\rangle=M$. In fact, the only way that to unit vectors $v$ and $w$ can satisfy this equation is to have $Mw=Tv$. (Since we know that $\|w\|=1$ and $\|Tv\|\leq M$, the Cauchy-Schwarz inequality tells us that $|\langle w,Tv\rangle\|\leq M$, and the equality case is only attainable when $Tv$ is a scalar multiple of $w$, being $M$ the only possible value of the scalar.)

But by selfadjointness of $T$, we also know that $\langle v,Tw\rangle=M$, so that $Mv=Tw$.

Now, one of the two vectors $v\pm w$ is nonzero, and we can compute

$T(v\pm w)=Tv\pm Tw=Mw\pm Mv=M(w\pm v)=\pm M(v\pm w)$.

This concludes the proof that $\pm\|T\|$ is eigenvalue with eigenvector $v\pm w$. The reality of the other eigenvalues can be proved by induction, restricting to $(v\pm w)^\bot$ as in the usual proof of the spectral theorem.

Remark: The proof above works with real or complex spaces, and also for compact operators in Hilbert spaces.

Comment: I would like to know if this proof can be found in the literature. I obtained it while trying to simplify a proof of the fact that if $T$ is a bounded selfadjoint operator, then $\|T\|=\sup_{\|v\|\leq 1} \langle Tv,v\rangle$ (as found, for example, on p.32 of Conway J.B., "An Introduction to Functional Analysis"). In the case of non-compact operators, one can only prove that $T$ has as an approximate eigenvalue one of the numbers $\pm\|T\|$. The argument is similar to the one above, but knowledge of the equality case of Cauchy-Schwarz is not enough. One has to know that near-equality implies near-dependence. More precisely, let $v$ be a fixed unit vector, $M\geq 0$ and $\varepsilon\in[0,M]$. If $z$ is a vector with $\|z\|\leq M$ such that $|\langle v,z\rangle|\geq \sqrt{M^2-\epsilon^2}$, then it can be proved that $z$ is within distance $\varepsilon$ of $\langle v,z\rangle v$.

Answer by Marcos Cossarini for Relating the angle between two vectors to max and min eigenvalues

2012-12-30T02:29:18Z

Problem: Let $T$ be a positive definite selfadjoint operator in an $n$-dimensional inner product space $H$. Find the maximum possible angle between a vector $v\neq 0$ and its image $Tv$, expressed in terms of the eigenvalues $\theta_0^2\geq\dots\geq\theta_{n-1}^2>0$ of $T$.

Solution: We want to minimize $\cos(v,Tv)=\frac{\langle v,Tv \rangle}{\|v\|\|Tv\|}$, or equivalently, its square $\frac{\langle v,Tv\rangle^2}{\langle v,v\rangle\langle Tv,Tv\rangle}$.

Since the angle between any vector $v\neq 0$ and its image $Tv$ doesn't change if we rescale $v$, we can rescale at our wish. I choose to restrict to those $v$ such that the numerator $g(v)=\langle v,Tv\rangle$ equals 1, so now the problem is equivalent to minimizing the denominator $f(v)=\langle v,v\rangle\langle Tv,Tv\rangle$, and now there are no divisions bothering.

Critical points of this restricted function are given by the equation $df(v)=\lambda dg(v)$, where $\lambda\in\mathbb R$ is a Lagrange multiplier. We calculate

$dg(v)=\langle v,T-\rangle+\langle -,Tv\rangle=2\langle Tv,-\rangle$ and

$df(v)=2\langle v,-\rangle\langle Tv,v\rangle+2\langle v,v\rangle\langle Tv,T-\rangle=2\|Tv\|^2\langle v,-\rangle+2\|v\|^2\langle T^2v,-\rangle$.

The critical point equation can now be rewritten:

$\langle \|Tv\|^2v+\|v\|^2T^2v-\lambda Tv,-\rangle=0$

and this is true iff $\|Tv\|^2v+\|v\|^2T^2v-\lambda Tv=0$. So $v$, $Tv$, and $T^2v$ are linearly dependent when $v$ is a critical point. But we know by Vandermonde [...] that they would be independent if $v$ had nonzero projections in three eigenspaces.

So the critical points are found in the planes spanned by two eigenvectors, and then we must solve our problem for $n=2$, being only interesting the case in which the two eigenvalues are different, because otherwise $v$ will also be an eigenvector and then the angle will be zero.

Calculations seem to get a little nicer if we express $T=S^2$, by letting $S$ be the only positive selfadjoint square root of $T$, with eigenvalues $\theta_0\geq\dots\theta_{n-1}>0$. We then have $g(v)=\langle Sv,Sv\rangle=\|Sv\|^2$, and $f(v)=\|v\|^2\|S^2v\|^2$.

Also, the symmetry is greater if we work in terms of the variable $w=Sv$, so that $g=\|w\|^2$ and $f=\|S^{-1}w\|^2\|Sw\|^2$ (this is not very important). If we express $w=x v_i+ y v_j$, where $v_i$ and $v_j$ are eigenvectors of $S$ with eigenvalues $\theta_i$ and $\theta_j$ and $i>j$, the Lagrange multipliers equation can be solved after some calculations [...], finding four critical points $(x,y)=(\pm \sqrt{\frac 12},\pm \sqrt{\frac 12})$. The minimum cosine for $w$ in the plane spanned by $v_i$ and $v_j$ is then $\frac{\langle w,w\rangle}{\|S^-1w\|\|Sw\|}=\dots=\frac{\theta_i\theta_j}{2(\theta_i^2\theta_j^2)}=\frac 12 (\frac{\theta_i}{\theta_j}+\frac{\theta_j}{\theta_i})$.

Once the case $n=2$ is solved, we want to select the plane so that $(\frac{\theta_i}{\theta_j}+\frac{\theta_j}{\theta_i})$ is maximum. But the function $h(t)=t+t^{-1}$ increases in the interval $[1,+\infty)$, so the value $t=\frac{\theta_i}{\theta_j}$ should be chosen as large as possible by maximizing $\theta_i$ and minimizing $\theta_j$.

The dots "..." represent explanations that I'm omitting. Request further details as needed.

ADDED by request of the OP: The "Vandermonde principle" says that an $n\times n$ Vandermonde matrix with different rows is invertible. This implies that if a vector $v$ has nonzero components in the eigenspaces corresponding to three different eigenvalues $\lambda_0$, $\lambda_1$, $\lambda_2$, then $v$, $Tv$ and $T^2v$ are independent.

Proof: Write $v=v_0+v_1+v_2$ so that for each $0\leq i<3$ we have $v_i$ a nonzero eigenvector corresponding to eigenvalue $\lambda_i$. Then we also have

$Tv=\lambda_0v_0+\lambda_1v_1+\lambda_2v_2$ and

$T^2v=\lambda_0^2+\lambda_1^2v_2+\lambda_2^2v_2$.

Our vectors $v$, $Tv$ and $T^2v$ belong to the subspace $S$ with basis $B=(v_0,v_1,v_2)$. And the coordinates of $v$, $Tv$ and $T^2v$ in basis $B$ are

$[v]_B=\left(\begin{array}{c}1\\1\\1\end{array}\right)$

$[Tv]_B=\left(\begin{array}{c}\lambda_0\\ \lambda_1\\ \lambda_2\end{array}\right)$

$[T^2v]_B=\left(\begin{array}{c}\lambda_0^2\\ \lambda_1^2\\ \lambda_2^2\end{array}\right)$

These are the columns of a Vandermonde matrix, so they are linearly independent, and so must be the vectors $v$, $Tv$, $T^2v$.

Answer by Marcos Cossarini for Intuition for mean curvature.

2012-12-29T22:44:12Z

Let $M$ be an oriented hypersurface in an oriented euclidean space $E$.

The normal curvature at a point $x_0\in M$ measures how much it is displaced, in the positive normal direction, with respect to the points of $M$ that are near it, in the same way that the Laplacian of a function measures how much the value of a function at a point differs from the average of the points that are near.

In fact, if $M$ is the graph of a nearly constant function $f:\mathbb R^n\to\mathbb R$ (for example, the vibrating membrane of a drum at some instant) then the Laplacian of $f$ is a good approximation of the normal curvature of $M$.

More generally, if $M$ is possibly not nearly flat, then for any $x_0\in M$ you can find the hyperplane $\Pi$ that is tangent to $M$ at $x_0$, and find a parametrization $f:\Pi\to M$ (only defined on a neighbourhood of $x_0$) such that $p\circ f=id_\Pi$, where $p:E\to\Pi$ is the orthogonal projection. This parametrization shows that $M$ is locally the graph of a function $g=q\circ p$ (where $q$ is the projection on the orthogonal complement of $\Pi$) such that $dg(x_0)=0$. Then, the normal curvature of $M$ at $x_0$ equals $\Delta g(x_0)$.

Answer by Marcos Cossarini for Does every ellipse inside a tetrahedron inside a ball fit in a triangle inside the ball?

2012-10-03T21:07:38Z

Some remarks (2012/10/3):

(0) If the statement is true, then it is tight in the following example:

Draw a circle A, that represents a slice of the ball.

Inscribe inside A a square B, which represents a very thin tetrahedron.

Inscribe inside B a square C joining the midpoints of the sides of B. This square represents the midbase of the tetrahedron.

Inscribe inside C a circle D, which is the ellipsoid.

In this case we have concentric circles with radius(A)=2radius(D), whicch is just enough to fit a triangle between A and D.

This example makes the problem beautiful for me.

(1) If the statement is true, then it is true also for the case in which the outer ball is generalised to an ellipsoid. Why? because by changing the inner product of the space, the ellipsoid turns into a ball.

(2) So the context of our problem is affine geometry of $\mathbb R^3$ (that is, we can drop the inner product). In fact, we can drop even the affine structure and keep only the projective structure.

(3) We can then settle a new inner product so that the ellipse is a circle.

More remarks (2012/10/4):

(4) We have an ellipse inside another one (the intersection between $B$ and the plane of the ellipse, and we are trying to fit a triangle between them.

The problem of finding, given two conic sections, finding an $n$-side polygon circumscribed to the inner conic section and inscribed in the outer conic section is called "Poncelet's second problem" (Santaló, "Geometría proyectiva", p.243-245).

Remarkably, if there exists a triangle that fits between both conic sections, then it is easy to find it: One can choose any point on the outer conic as vertex of the triangle, and the construction will work. See "Poncelet's porism" at http://sbseminar.wordpress.com/2007/07/16/poncelets-porism/ .

I will say that a conic $n$-fits inside another one iff an $n$-sided polygon can be fit between them.

(5) By doing some projective transformations, I think the problem can be reduced to proving the following: Let $0\le a\le b\le 1$. Let $B$ be the cylinder $x^2+y^2\le 1$, $E$ the ellipse in the plane $z=0$ given by $(\frac xa)^2+(\frac yb)^2=1$. If there is a tetrahedron $T$ such that $E\subseteq T\subseteq B$, then there is a triangle $T'$ in the plane $z=0$ such that $E\subseteq T'\subseteq B$.

I don't have a complete proof of this reduction, but I can give more details.

(6) In the above situation, the existence of the triangle $T'$ is equivalent to the fact $a+b\leq 1$. This can be observed by trying to construct a triangle starting with a vertex in the $x$ axis. The starting point is irrelevant by remark (4).

(7) By remarks (5) and (6), we have to prove that if a tetrahedron fits between the cylinder and the ellipse, then $a+b\leq 1$. To fully flatten the problem, we have to be able to recognise if a given quadrilateral surrounding our ellipse is the projection of a tetrahedron that surrounds the ellipse. If we are given the projection $ABCD$ of the vertexes of the tetrahedron, and the quadrilateral $PQRS$ where $T$ intersects the plane $z=0$ (with $P\in[A,B]$, $Q\in[B,C]$, etc), then we can see if the ellipse fits inside $PQRS$. But also, applying Ceva's theorem, it can be shown that $\frac{|P-A| |Q-B| |R-C| |S-D|}{|P-B| |Q-C| |R-D| |S-A|}=1$, and this equation can be used to confirm that the points $A,B,C,D,P,Q,R,S$ of the plane $z=0$ where indeed obtained from a tetrahedron by projecting on and intresecting with the plane $z=0$.

(8) Some experiments that I did with the software GeoGebra suggest that an ellipse $A$ $3$-fits inside another ellipse $C$ iff there is an intermediate ellipse $B$ such that $A$ 4-fits inside $B$ and $B$ 4-fits inside $C$. I think that there is a path of ellipses joining $A$ and $C$, but to define it I would need a notion of $n$-fitting with $n$ non integer.

Answer by Marcos Cossarini for Optic fibers after Joseph O'Rourke

2012-04-12T21:33:53Z

I don't know why this question appeared yesterday in my main mathoverflow screen, since the last comment appears to be from March 1st. I also think the conjecture is true, but my argument has plenty of holes. Here's what I thought.

Let $\Omega$ be an optic fiber that is the interior region of a compact connected $\mathcal C^1$ surface in $\mathbb R^3$, that is the union of a reflective surface $R$, a starting surface $F_0$ and an ending surface $F_1$. I assume (0) all optic fibers are like this.

Assume (1) that you can foliate $\Omega$ by a uniparametric family of plane surfaces $(F_s)_{s\in[0,1]}$, such that every ray starting at $F_0$ with positive initial $\dot s$ keeps a positive $\dot s$, until it reaches $F_1$. Let's call this a foliated optic fiber. Assume (2) that all the $F_s$ are plane surfaces and call it planely foliated, and let's call the fiber Petruninean if each $F_s$ is orthogonal to the reflective surface $R$ along its border.

Claim (3): Every planely foliated optic fiber is Petruninean.

Let my say why I believe it.

Let $m$ be the last value such that the fiber is Petruninean for $s\in[0,m]$. By finding a foliation of curves that is orthogonal to $F$ we can identify the points of $F_s$ at different values of $s$, and think of $F_s$ as a uniparametric family of regions in $\mathbb R^2$.

Conjecture (4): $F_s$ has constant area.

I don't know why this should be true, but it seems to be clear to Anton (if I understood well his comment concerning the Liouville theorem), and I would like to know why. Is it evident after studying billiard problems? I'll assume it true.

Can $F_s$ be nonconstant? It is constant until $s=m$, and then it starts to change. Since the area is conserved, it must advance at some points of its border and recede at other points. Find a point where it has just started receding. It represents a point $P\in R$. Fire a ray from $P$ into $\Omega$, with initial velocity orthogonal to $R$. Because of how we chose $P$, it will have negative initial $\dot s$, and by adjusting our choice of $P$ I would like (5) to be able to ensure that it reaches $F_m$. After that the optical fiber is Petruninean, and the ray makes its way to $F_0$. By reversing this ray, we obtain a ray that starts at $F_0$, reaches $P$, and bounces back to $F_0$, so $\Omega$ was not an optic fiber, after all.

The holes in the argument suggest more questions:

Question 0: Is every foliated optic fiber planely foliated?

If $\Omega$ is foliated, the phase space is the disjoint union of $T\Omega_+={\dot s>0}$, $T\Omega_-{\dot s<0}$, and their zero-measure common border $T\Omega_0={\dot s=0}$. Does every phase in $T\Omega_+$ correspond to a ray that came from $F_0$ and is going to $F_1$? If this is true (and also the analogous statement for $T\Omega_-$), then it is easy to see (by shooting rays between points of the same $F_s$) that every $F_s$ must be locally convex, and hence plane.

Question 1: Is every optic fiber foliated?

We can express the phase space as a union of the set $T\Omega_+$ of phases corresponding to rays that go from $F_0$ to $F_1$, the set $T\Omega_-$ defined analogously (so that $\dot x\in T\Omega_+$ iff $-\dot x\in T\Omega_-$), and the set $T\Omega_0$ of phases corresponding to rays that remain inside $\Omega$ eternally (in both directions of time). Observe how for each $x\in\Omega$, the classification of phases partitions $T_x\Omega$ into two opposite cones $T_x\Omega_+$, $T_x\Omega_-$ and a selfopposite cone $T_x\Omega$.

Does $T\Omega_0$ have measure zero? Are the cones convex? I have no idea.

Can we totally order the phases of $T\Omega_+$, so that it is then possible to assign them a scalar parameter, that makes possible to apply supremum arguments to prove things? At least we can partially order it. Do supremum arguments work in posets?

Answer by Marcos Cossarini for Why are so few operations with arity bigger than 2?

2012-02-05T20:40:25Z

The composition law in a monoid is usually represented using a binary operation (multiplication) and a zeroary operation (unit), but I view it more naturally as an operation (say, a bracket) taking any finite list of arguments and being associative in the sense that brackets can be eliminated in any expression, for example we have the identities

[a,[],b,[[c,d],e],[f]]=[a,b,c,d,e,f]

and

[a]=a.

Then we can define a zeroary operation 1:=[] and a binary operation a*b:=[a,b], and recover the bracket from them, using identities such as [a,b,c,d]=[a,[b,[c,d]]]. These two operations satisfy the usual axioms of a monoid, and any two operations satisfying them can be extended to an associative finitary bracket.

I view the usual representation by a binary and a zeroary operation as an artifact for being able to produce simpler-looking proofs that the structures that we encounter are monoids.

My point is that naturally binary operations are not that common either! Perhaps an example is the Lie bracket.

Answer by Marcos Cossarini for The "Dzhanibekov effect" - an exercise in mechanics or fiction? Explain mathematically a video from a space station

2011-11-27T05:23:07Z

I started adding comments to Victor's answer, but they finally outgrew to this:

From an elementary point of view, the energy is conserved, and also the angular momentum vector $L$ (measured with respect to an inertial frame).

From a Lagrangian-mechanics-on-manifolds point of view, we have to solve a problem of movement in $SO_3$, that can be parametrized using Euler angles. Since there is only kinetic energy (no potential), it is equivalent to a problem of geodesic flow in a Riemannian manifold, and affine reparametrizations of solutions will again be solutions. This simply means that every solution can be sped up at wish.

Only the first angle in the composition will be a cyclic coordinate, but this is an artifact of the parametrization, since we see that there is $SO_3$ symmetry, i.e., the whole "position component" should be cyclic, and we would expect to reduce the second order o.d.e. on "position" to a (first order) problem on "speed".

(Is this intuition correct? Can it be done in other problems?)

To do this, we go to the frame of reference of the body. We try to find the motion of the angular speed vector $\omega$ with respect to it . This is a first order three dimensional problem, governed by the Euler equations $I_1\dot\omega_1=(I_2-I_3)\omega_2\omega_3$ (1)

$I_2\dot\omega_2=(I_3-I_1)\omega_3\omega_1$ (2)

$I_3\dot\omega_3=(I_1-I_2)\omega_1\omega_2$ (3),

which are expressed in principal axes coordinates.

(I understand this way of changing reference frames as an abuse of notation: the two reference frames $body$ and $space$ are different (Euclidean) spaces (even before coordinatizing them), and they are related by an isometry $body\to space$ (the attitude of the body) that changes with time (according to a function that we are trying to find). So every vector is two vectors: we actually have two angular speed vectors $\omega_{space}$ and $\omega_{body}$. This is relevant when we derivate them with respect to time. For example, $L_{space}$ is constant but $L_{body}$ isn't. This idea is simply a coordinate-free formalization of the notion of measuring speed "with respect to a reference frame".)

We can also express the problem in terms of the angular moment vector $L$, which is linearly equivalent to $\omega$ via the equation $L=I\omega$. I will switch between both versions when convenient.

Using the conservation of energy $2E=\langle\omega,I\omega\rangle$, we see the movement is constrained to an ellipsoid. So we have a two dimensional first order problem. You can search "Poinsot ellipsoid" in Google Images to see it with the flow lines drawn on it. The only equilibrium points are (from looking at the Euler equations) at the principal axes of the ellipsoid, which are the lines along which uniform rotation is possible.

Assuming $I_1>I_2>I_3$ we find only six equilibrium points, for example: $(\sqrt{2E/I_1},0,0)$. Near it we can express $\omega_1=\sqrt{(2E-I_2\omega_2^2-I_3\omega_3^2)/I_1}$ and write the second and third Euler equations for $\omega_2$ and $\omega_3$. Linearizing at $(0,0)$ we get

$I_2\dot\omega_2=(I_3-I_1)\sqrt{2E/I_1}\omega_3$ (2')

$I_3\dot\omega_3=(I_1-I_2)\sqrt{2E/I_1}\omega_2$ (3')

and since $I_3-I_1<0$ and $I_1-I_2>0$, we get imaginary eigenvalues, which means the equilibrium point is possibly neutrally stable. We confirm that the trajectories around it are closed by observing the reflection symmetry $(\omega_2,\omega_3)\to(-\omega_2,\omega_3)$. This happens at the equilibriums associated to the smallest moment of inertia $I_3$. At $(0,\sqrt{2E/I_2},0,0)$ we have two real opposite eigenvalues, which means the equilibrium point is unstable. We can determine the topology of the flow lines completely by analyzing the signs in the Euler equation in each octant, but it is easier to use another constant of movement.

The length of $L$ is constant, so the trajectory will be the intersection of the Poinsot ellipsoid with a sphere (or with an ellipsoid having the same principal axes, to phrase everything in terms of $\omega$ instead of $L$, but what follows is most clearly seen in the $L$ picture).

If the sphere is smallest, we get a pair of opposite points on the first axis.

If the sphere is small, we get a pair of closed curves around the first axis.

If the sphere is large, we get a pair of closed curves around the third axis.

If the sphere is largest, we get a pair of points on the third axis.

The middle case, corresponding to a medium sphere, contains the equilibrium points of the second axis (lets call them $A$ and $B$), joined by four curves (lets call them special). Each special curve is topologically equivalent to an open interval, and travel along it requires infinite time (since from unicity of solutions, we can't expect to reach an equilibrium point in finite time).

If we attempt to make a body rotate along an unstable axis in practice, we will miss, and obtain a closed curve around the first or third axis that will pass periodically near $A$ and $B$. Since motion near an equilibrium point is slow, the curve will spend most time near them, and will be similar to a pair of special curves. If we speed this up by giving a high initial energy, we get this periodic, almost instantaneous switch from $A$ to $B$.

With respect to an inertial frame, since $L_{space}$ is constant, we see that the body always rotates in the same direction, switching its position to the opposite after a period $T$. The movement is quasi-periodic: after time $T$, the movement restarts but with a possible angular shift around the $L_{space}$ axis (whose magnitude is always the same for a certain trajectory and can't be calculated easily (as far as I know)).

Matrix representation of real *-algebras

2011-06-03T21:40:43Z

It is a standard fact that everyt real $n$-dimensional algebra is a subalgebra of $M_n(\mathbb R)$.

The transposition map, operating in $M_n(\mathbb R)$, is an involutive ($(A^t)^t=A$) antiautomorphism (an $\mathbb R$-linear isomorphism satisfying $(AB)^t=B^tA^t$). This makes $M_n(\mathbb R)$ a real *-algebra. The same is true in every transpose-closed subalgebra of $M_n(\mathbb R)$.

Is every real *-algebra of this kind? (a transpose-closed subalgebra of $M_n(\mathbb R)$, with the * represented by transposition)

This is true at least in the famous real *-algebras $\mathbb C$ and $\mathbb H$.

Towards a metric characterization of Euclidean spaces

2011-05-08T08:42:49Z

I want to obtain a metric characterization of the classical finite dimensional spaces of Euclidean geometry.

Motivation: Suppose $A$ and $B$ live in an $n$-dimensional Euclidean space. They are each assigned the task of constructing an equilateral triangle of side length 5. Subject $A$ finds first one point $p_1$. If $n>0$, he can then find another point $p_2$ at distance 5 from $p_1$. If $n>1$, he can then find $p_3$ at distance 5 from $p_1$ and $p_2$. Then $B$ proceeds similarly. He selects a point $q_1$ (probably different from $p_1$), then finds another point $q_2$ and $q_3$. We wouldn't expect him to get stuck before constructing $q_3$, since $n$ is the same for both subjects. I will say a space is all-equal if the possibility of completing a figure is independent of the starting points chosen. Are Euclidean spaces all-equal? Are there non Euclidean all-equal spaces?

Precise statement

Definition: A metric space $X$ is all-equal if every time $S$ and $T$ are isometric subspaces of $X$ (with a selected isometry $S\to T$), the isometry extends to an automorphism of $X$.

Question 0: Is every Euclidean space all-equal?

To state the second question we must first observe some limitations. Being connected our life intervals, we cannot visit more than one connected component of our space, and being imprecise our measurements, we cannot distinguish directly between a space and its completion. Finally, existing no natural unit of measurement, the correct category to state these questions is that of spaces in which the distance is defined up to scale. That is, it takes values in a 1-dimensional module $M$ over $[0,+\infty)$, with no multiplication structure (although lengths can be tensorised to obtain areas).

More definitions: A congruence space is a pair $(X,M,d)$ where $X$ is a set, $M$ is a 1-dimensional $[0,+\infty)$ module and d is a distance in $X$ taking values in $M$. The morphisms from $(X,M,d)$ to $(Y,N,e)$ will be the subsimilarities, that is, the pairs $(f,\alpha)$ where $f$ is a function from $X$ to $Y$ and $\alpha$ is a morphism from $M$ to $N$ such that for each $x$, $x'$ in $X$ we have $e(fx,fx')\leq\alpha(d(x,x'))$. Similarities are similarly defined using an $=$ sign instead of $\leq$. Subspaces are defined by restriction, and the identity of $(X,M,d)$ is the obvious $(id_X,id_M)$. It's easy to prove that every isomorphism is a similarity. A congruence space $X$ is all-equal if every time $S$ and $T$ are similar subspaces, the similarity extends to an automorphism of $X$.

Question 1: Is every connected complete all-equal space a finite dimensional real inner-product space?

Remarks: if the connectedness hypothesis is dropped, some discrete spaces would be all-equal. If we work in the usual category of metric spaces, projective and hyperbolic spaces could be all-equal, and also Euclidean spheres, both with their inner metric and with the subspace metric.

Partial results and lines of thought:

EDIT: As noted by Sergei, there are counterexamples. Generally, if $(X,d)$ is an all-equal metric space, and $p>1$, then $(X,d^{\frac 1p})$ could be an all-equal space. This is similar to the example in which However, I was thinking of spaces in which the distance from $x$ to $y$ is measured using a signal that travels form $x$ to $y$ at constant speed. Hence I require that the space have the following property:

If $x$ is a point and $a$ and $b$ are lengths, then $N_b(N_a({x}))=N_{a+b}({x})$ (where the $N$ stands for open neighborhood of a set).

Related MO questions: http://mathoverflow.net/questions/47882/characterizations-of-euclidean-space http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean http://mathoverflow.net/questions/8513/characterization-of-riemannian-metrics

Answer by Marcos Cossarini for Towards a metric characterization of Euclidean spaces

2011-05-08T20:52:47Z

Since the post is getting too long, I will place here my partial results.

If a real Banach space is all-equal, I think that it is finite dimensional and inner-product. This can be proved (at least in the finite dimensional case) using the John ellipsoid (see http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean), even though I don't like to introduce such a cumbersome structure.

If a Riemannian manifold is all-equal, then it is homogeneous and with constant curvature (how do we deal with the fundamental group?). If a Finsler manifold is all-equal, then I would expect it to be Riemannian (again using the John ellipsoid). I have no idea how to proceed without differential structure.

The hypothesis added after edition is equivalent to saying that if $x$, $z$ are points and $a$, $b$, are lengths such that $a+b>d(x,z)$, then there exists $y$ such that $a>d(x,y)$ and $b>d(y,z)$.

Now it is easy to construct approximate midpoints, then midpoints (using completeness), then segments, then complete geodesics (a segment can be extended by moving the starting point to the midpoint and the midpoint to the end (EDIT:the starting point must go to the starting point, and the midpoint to the end)). Then every pair of points can be joined by a geodesic of the expected length.

Are geodesics unique, or is it possible that there is an ugly graph embedded in the space? (See http://mathoverflow.net/questions/12394/representability-of-finite-metric-spaces, an example of ugly graph is a 4-cycle with the path distance.)

Is it possible to define angles? That is, suppose $S$ and $T$ are rays with common origin $x$. If I consider $s$ and $t$ their respective intersections with a ball $B_r(x)$ and $s'$, $t'$ their respective intersections with the ball of radius $B_{r'}(x)$, is it true that $\frac{d(s,t)}{r}=\frac{d(s',t')}{r'}$?

Actually, since every sphere in an all-equal congruence space is all-equal (as a metric space; it doesn't have dilations), it would be nice to define an intrinsic metric in each sphere, so that the angle between $S$ and $T$ could be defined as $\frac{i(s,t)}{r}$, where $i$ is the intrisic metric of $B_r(x)$.

Answer by Marcos Cossarini for Which norms have rich isometry groups?

2011-05-10T03:34:37Z

As implicit in Bill's answer and prooved in Konrad's the answer to http://mathoverflow.net/questions/13596/continuous-automorphism-groups-of-normed-vector-spaces, any isometry group of a normed space is a subgroup $G$ of $O(n)$ that contains $-id$.

It may be a good starting point.

Does the positive dimension imply that the space contains an Euclidean plane?

Forget the original norm and consider Euclidean $n$-space with a closed subgroup of $O(n)$ acting on it.

Let $A$ be any non zero element of the Lie algebra tangent to $G$. Since $exp(\epsilon A)$ is a rotation, and every rotation is equivalent to a matrix that has $2\times 2$ rotation blocks in the diagonal and zeros elsewhere. Expressed in the same (or other?) coordinates, $A$ probably had $2\times 2$ antisymmetric blocks in the diagonal and zeros elsewhere (although I don't know right now how to prove this fact without abandoning the algebra). The point having first coordinate $1$ and then zeros moves in a circle around the origin. Hence the spaces contains an Euclidean plane.

Related stuff:

http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean http://mathoverflow.net/questions/12452/maximal-ellipsoid http://mathoverflow.net/questions/64269/towards-a-metric-characterization-of-euclidean-spaces (not too related but I'm trying to promote it).

Is there a category for pre-smooth spacetime geometry?

2011-05-08T08:52:46Z

I start with an Euclidean space. I drop global symmetry and I get a Riemannian manifold. I drop symmetry (isotropy) in the tangent space and I get a Finsler manifold. I drop differentiability and I get a metric space.

What happens if I start with Minkowski's spacetime? I drop global symmetry and I get a pseudo-Riemannian manifold. Can I drop symmetry of the tangent spaces? (Question) Can I drop differentiability? (Main question)

Has anyone studied categories analog to that of metric spaces or Finsler manifolds, but including a notion of causality/order/passage of time?

Answer by Marcos Cossarini for Find symmetries of a tree

2010-05-03T07:17:45Z

I don't understand how the angles of the connecting finite segments are determined, so I'll assume the angles are set so that they don't break any symmetry. First observe that the reflection wrt the real axis sends sectors 0,1,2,3,4,5 to 0,5,4,3,2,1 respectively. So in your second example, tree

(0,1,5)(1,2,5)(2,4,5)(2,3,4) turns into

(0,5,1)(5,4,1)(4,2,1)(4,3,2)

which is different from the original tree (the original and transformed tree share only the first and last junction). So the transformation is not a symmetry of the tree. However, the same transformation sends the first example

(0,1,5)(1,2,4,5)(2,3,4) to

(0,5,1)(5,4,2,1)(4,3,2)

which is the same tree, represented in a non standard way because the junctions appear in the wrong order and the sectors of each junction are also in the wrong order.

Rotation of 180 degrees sends 0,1,2,3,4,5 to 3,4,5,0,1,2 (add 3 mod 6) so

(0,1,5)(1,2,5)(2,4,5)(2,3,4) turns into

(3,4,2)(4,5,2)(5,1,2)(5,0,1)

which is the same tree, again represented in a non standard way (the junctions appear in the inverse order, and each junction has its inciding sectors cycled).

So the recipe seems to be the following: Find out, for your transformation of the plane, which sectors goes to which, apply this permutation to the tree representation, and then reorder each tree representation (original and transformed) in a standard way that allows to compare if they are equal. If they are equal, then (assuming the angles are nice), the transformation of the plane is a symmetry of the tree. If they are not equal, then the transformation is not a symmetry of the tree.

Answer by Marcos Cossarini for Why worry about the axiom of choice?

2010-05-02T22:04:37Z

As Pete Clark says,

"In general, I think "Why do people worry about X?" is primarily psychological. I can (and alas, do) worry about the correctness of my proofs for lots of reasons...but fundamentally I am worried about myself and not about some mathematical object or principle."

Let me explain why I worry about the axiom of choice and, generally, about non-constructive methods in mathematics. One of the fascinating aspects of mathematical activity is that it produces results that can trascend our life, like the Pythagorean theorem or Newton's binomial formula.

Many concepts of (formalized) mathematics evolved as ideal models of concrete things: some functions can model the movement of a particle, Turing machines model something following an algorithm, etc.. Althought the relationship between the abstract, formal concept and the "real" thing is sometimes loose, it is a historical fact that the concepts were devised as models of such concrete things.

Mathematical theorems proved in a constructive way generally have a close connection to algorithmic processes that can possibly be relevant for humans of all times. But for example, the axiom of choice establishes a similarity between finite and infinite sets (which are qualitatively different, for example, no finite set is bijective to a propert subset) that is difficult to justify from an objective point of view. It states that infinite products of nonempty sets are nonempty. As far as I see, there isn't an objective argument to prefer the axiom of choice to it's negation. Future generations of mathematicians could possibly negate it (or more provably, live it undecidable) and a lot of parts of twentieth century mathematics would cease being theorems.

However, it seems less likely that a future mathematical language would cease providing concepts that mimick, for example, automata.

Note: The indecidability of the axiom of choice in ZF theory is still a metatheorem, so if a ZF statement can be proved false using AC, then it is (meta) clearly not a theorem. So the axiom of choice could still be useful for studying ZF, as a tool to produce counterexamples that limit our theorem-proving pretensions, but with the same formal status as, say, its negation.

Also note: I see that this answer is not part of math if math is understood as the study of the consequences of the ZFC axioms, and also it doesn't deal with the specific questions, but I hope that it explains why (some) people worry and fuss about the axiom of choice. It may be relevant, for example, for math professors, that could atract a wider audience by presenting a more secular version of mathematics that excludes the axiom of choice from the set-theoretic formalism. Indepently of weather they use it or not in their own work.

Answer by Marcos Cossarini for Examples of mathematics motivated by technological considerations

2010-03-10T02:59:38Z

One example of mathematical achievement that has application to the "industry" of medicine is the Radon transform, that enables to produce x-ray tomographies.

The Radon transform of a plane section of your body is the set of information that you obtain after shooting x-rays through all the possible lines contained in that plane, and recording the intensity of the ray that comes out at the other side. More formally, its an integral transform: it's the result of integrating the density of the body tissue along each of the lines.

Johann Radon defined his transform in 1917, and calculated a formula for its inverse. That is, he deviced a way to recover from the transformed data the density of tissue in each point of the plane section. The inversion formula involved a lot of calculations that couldn't be handled at that time.

Cormack and Newbold implemented the idea with the aid of computers to handle the calculations. They got a Nobel Price in 1979 because of this.

Answer by Marcos Cossarini for Is every finite group a group of "symmetries"?

2010-02-20T16:14:07Z

The permutohedron may have additional symmetries. For example, the order 3 permutohedron {(1,2,3),(1,3,2),(2,1,3),(3,1,2),(3,2,1)} is a regular hexagon contained in the plane x+y+z=6, which has more than 6 symmetries.

I think we can solve it as follows:

Let G be a group with finite order n thought via Cayley's representation as a subgroup of S_n.

Let S={A_1,...,A_n} be the set of vertices of a regular simplex centered at the origin in a (n-1)-dimensional real inner product space V. Let r be the distance between the origin and A_1. The set of vertices S is an affine basis for V.

First unproven claim: If a closed ball that has radius r contains S, then it is centered at the origin. Let B be this ball.

The group of isometries that fix S hence contains only isometries that fix the origin and permute the vertices, which can be identified with S_n in the obvious way. The same is true if we replace S by its convex hull.

Now G can be thought of as a group containing some of the symmetries of S.

Let C=k(A_1+2A_2+3A_3+...+nA_n)/(1+2+...+n), with k a positive real that makes the distance between C and the origin a number r' slightly smaller than r.

Let GC={g(C) / g in G}. It has n distinct points, as a consequence of being S an affine basis of V.

Let P be the convex hull of the points of S union GC.

Remark: A closed ball of radius r contains P iff it is B. The intersection of the border of B and P is S.

Second unproven claim: The extremal points of P are the elements of S union GC.

Claim: G is the group of symmetries of P.

If g is in G, g is a symmetry of GB and of S, and it is therefore a symmetry of P. If T is a symmetry of P, then T(P)=P, and in particular, T(P) is contained in B, and hence T(0)=0 (i.e. T is also a symmetry of B). T must also fix the intersection of P and the border of B, so T permutes the points of S, and it can be thought of as an element s of S_n sending A_i to A_s(i). And since T fixes the set of extremal points of P, T also permutes GC. Let's see that s is in G.

Since T(C) must be an element of g(C) of GC, we have T(C)=g(C). But since T is linear, T(C/k)=g(C/k). Expanding,

(A_s(1)+2A_s(2)+...+nA_s(n)/(1+...+n)=(A_g(1)+2A_g(2)+...+nA_g(n))/(1+...+n).

For each i in {1,...,n} the coefficient that multiplyes A_i is s⁻1(i)/(1+...+n) in the left hand side and g⁻1(i) in the right hand side. It follows that s=g.

I think that, taking n into account, the ratio r'/r can be set to substantiate the second unproven claim. The first unproven claim may be a consequence of Jung's inequality.

EDIT: With the previous argument, we can represent a finite group of order n as the group of linear isometries of a certain polytope in an n-1 dimensional real inner product space.

Now, if a finite group G of linear isometries of an (n-1)-dimensional inner product space V is given, can we define a polytope that has G as its group of symmetries? Yes. I'll give a somehow informal proof.

Let G={g_1,...,g_m}. Let A={a_1,...,a_n} be the set of vertices of a regular n-simplex centered at the origin of V. Let S be the sphere centered at the origin that contains A, and let C be the closed ball. Notice that C is the only minimun closed ball contaiing A.

(Remark: The set A need not be a regular simplex. It may be any finite subset of S that intersects all the possible hemispheres of S. C will then still be only minimum closed ball containing it.)

Remark: An isometry of V is linear iff it fixes the origin.

Before proceeding, we need to be sure that the m copies of A obtained by making G act on it are disjoint. If that is not the case, our set A is useless but we can find a linear isometry T such that TA does de job. We consider the set M of all linear isometries with the usual operator metric, and look into it for an isometry T such that for all (g,a) and (h,b) distinct elements of GxA the equation g(Ta)=h(Tb) does not hold. Because each of the n*m(n*m-1) equations spoils a closed subset of M with empty interior(*), most of the choices of T will do.

Let K={ga/g in G, a in A}. We know that it has n*m points, which are contained in the sphere S. Now let e be a distance that is smaller than a quarter of any of the distances between different points of K. Now, around each vertex a=a_i of A make a drawing D_i. The drawing consists of a finite set of points of the sphere S, located near a (at a distance smaller than e). One of the points must be a itself, and the others (if any) should be apart from a and very near each other, so that a can be easily distinguished. Furthermore, for i=1 the drawing D_i must have no symmetries, i.e, there must be no linear isometries fixing D_1 other than the identity. For other values of i, we set D_i={a_i}. The union F of all the drawings contains A, but has no symmetries. Notice that each drawing has diameter less than 2*e,

Now let G act on F and let Q be the union of the m copies obtained. Q is a union of n*m drawings. Points of different drawings are separated by a distance larger than 2*e. Hence the drawings can be identified as the maximal subsets of Q having diameter less than 2*e. Also, the ball C can be identified as the only sphere with radius r containing Q. S can be identified as the border of C.

Let's prove that the set of symmetries of Q is G. It is obvious that each element of G is a symmetry. Let T be an isometry that fixes Q. It must fix S, so it must be linear. Also, it must permute the drawings. It must therefore send D_1 to some gD_i with g in G and 1<=i<=n. But i must be 1, because for other values of i, gD_i is a singleton. So we have TD_1=gD_1. Since D_1 has no nontrivial symmetries, T=g.

We have constructed a finite set Q with group of symmetries G. Q is not a polytope, but its convex hull is a polytope, and Q is the set of its extremal points.

(*) To show that for any (g,a) and (h,b) distinct elements of GxA the set of isometries T satisfying equation g(Ta)=h(Tb) has empty interior, we notice that if an isometry T satisfies the equation, any isometry T' with T'a=Ta and T'b=/=Tb must do (since h is injective). Such T' may be found very near T, provided dimV>2. The proof doesn't work for n=1 or 2, but these are just the easy cases.

Comment by Marcos Cossarini

2013-04-19T14:36:34Z

For the non-Hausdorff case, see ncatlab.org/nlab/show/exponential+law+for+spaces.

Comment by Marcos Cossarini

2013-03-08T19:45:36Z

Notice that this Arens' space is slightly different from the one given in Wikipedia (en.wikipedia.org/wiki/Arens%E2%80%93Fort_space), since in this space the set $\{c\}\cup\{a_{n,m}:n,m\in\omega\}$ is not open.

Comment by Marcos Cossarini

2013-01-24T21:14:05Z

@Ben, the index in the coordinate expression $\frac{\partial f}{\partial x^j}$ for the 1-form $df$ is clearly in the low position! In fact, this is the main reason that I see for having to put the indexes of the coordinates in the high position as we do, instead of doing everything in the opposite way, which would be better in some way: we could write $f=x_1^2+x_3$ instead of $f=(x^1)^2+x^3$.

Comment by Marcos Cossarini

2013-01-24T21:03:08Z

Regarding differential geometry: If $f:M\to\mathbb R$ is a smooth function on a manifold and $x:M\to\mathbb R^n$ is a chart, I prefer $\left(\frac{\partial f}{\partial x}\right)_j$ or $\left(\frac\partial{\partial x}\right)_j f$ (or even $\partial_j f$ if the choice of the particular chart is clear or irrelevant). Because the notation $\frac{\partial f}{\partial x^j}$ suggests that $\frac{\partial f}{\partial g}$ could be defined using only $g$, and in fact you need to know that you are restricting to the curve along which the other coordinates $x^i$ are constant.

Comment by Marcos Cossarini

2013-01-14T00:33:00Z

I think that the main difference is that Alexander extremises $x^tAx$ and I extremise $y^tAx$. That the two situations are not trivially equal is the subject of p.32 of Conway.

Comment by Marcos Cossarini

2013-01-13T23:56:08Z

Alexander, when you said that the minimum is an eigenvalue, did you mean to prove it by applying the Lagrange multiplier equation to the function $f(x)=x^tAx$ restricted to a level set of $g(x)=x^tx$, or did you have a different idea in mind?

Comment by Marcos Cossarini

2013-01-13T23:45:26Z

But Lagrange multipliers is, in my opinion, different from the argument above, which in fact was originally designed to deal with bounded operators, as explained in the comment. Can Lagrange multiplier be used to prove that $\pm\|T\|$ is an approximate eigenvalue of a bounded operator T? If not, is this enough to conclude that the proofs are different?

Comment by Marcos Cossarini

2013-01-13T23:35:51Z

I don't understand Alexander's answer. How do you prove that if $R(x)=\frac{x^tAx}{x^tx}$ is maximum, then $x$ is an eigenvector? I got nowhere by derivating $R$, and the only easy way that I see to complete his proof is to normalize $x$ to get a maximum of $x^tAx$ in the unit sphere, and then write the Lagrange multipliers equation that tells you that $x$ is an eigenvector.

Comment by Marcos Cossarini

2013-01-13T14:21:39Z

I meant $\frac 12\sum a_{ij}^2$.

Comment by Marcos Cossarini

2013-01-13T02:00:10Z

To add a little more detail: The total energy $\frac 12\sum a_{ij}$, which is the sum of the energy on the diagonal and $\Sigma$, is invariant by orthogonal conjugation, so we want to move it to the diagonal. When you apply a rotation $J$ in the plane spanned by the canonic vectors $e_i$ and $e_j$, which only affects the $i$th and $j$th rows and columns, the resulting coefficients $ii$, $ij$, $ji$, $jj$ of $J^tAJ$ depend only on the same coefficients of $A$, so the problem is reduced to increasing the energy on the diagonal of a $2\times 2$ matrix.

Comment by Marcos Cossarini

2012-12-31T05:27:45Z

Not an answer, but see Caratheodory, "The most general transformations of plane regions which transform circles into circles". He doesn't require continuity, but does require bijectivity in some region. Haven't read it, though. I was redirected to there by McCallum, "GENERALIZATIONS OF THE FUNDAMENTAL THEOREM OF PROJECTIVE GEOMETRY". Also see Höfer, "A CHARACTERIZATION OF MOBIUS TRANSFORMATIONS", which deals with arbitrary dimension, but still requires injectivity in a region and that the map sends hyperspheres to hyperspheres.

Comment by Marcos Cossarini

2012-10-09T01:01:17Z

Good job! I don't understand the projection through $C$ from $\omega$ to $\Omega$ (in the comment between brackets in the last paragraph), but anyway, from the equality of ratios, it is easy to see that the fourth point of intersection $C′$ should satisfy $(C'A,C'C;C'X,C'Y)=(C'M,C'C,C'X,C'Y)$, so it should be on the line $AM$. In case anyone is wondering, the expression "cross ratio of the four points $A$, $C$, $X$, $Y$ with respect to $\omega$" is justified because given four points E,F,G,H and a number $\lambda$, the locus of the points T such that $(TE:TF:TG:TH)=\lambda$ is a conic section.

Comment by Marcos Cossarini

2012-10-05T03:59:51Z

Please add the tag "projective-geometry".

Comment by Marcos Cossarini

2012-10-04T23:46:37Z

Right now I have the following question: Assume we have an ellipse $E$ fitting tightly inside a tetrahedron $T$ (i.e., there is no tetrahedron containing $E$ that is strictly contained in $T$) so that $T$ fits tightly inside $B$ (i.e., there is no tetrahedron contained in $B$ that strictly contains $T$). Is it possible that we find another tetrahedron contained in $B$ and containing $E$ for which the fit is not tight?

Comment by Marcos Cossarini

2012-05-10T23:13:15Z

@Hans: Then you googled ""skipping rope equation"".