User leonard - MathOverflow

On the definition of ‘smooth vectors’ in Rieffel's “Deformation Quantization for Actions of $ \mathbb{R}^{d} $”.

2012-12-22T19:39:11Z

On the first page of Chapter 1 of Rieffel's Deformation Quantization for Actions of $ \mathbb{R}^{d} $, Rieffel defines a family of seminorms on the space $ A^{\infty} $ of smooth vectors of a Fréchet space $ A $, for some action $ \alpha $ of the Lie group $ \mathbb{R}^{d} $ on $ A $, as follows. Suppose we already have a family $ (\| \cdot \|_{i})_{i \in \mathbb{N}} $ of seminorms on $ A $ that determine its topology. Choose a basis $ \lbrace X_{1},\ldots,X_{d} \rbrace $ of $ \mathbb{R}^{d} $. Then for each $ k \in \lbrace 1,\ldots,d \rbrace $, let $ \alpha_{X_{k}} $ denote the operator of partial differentiation on $ A^{\infty} $ in the direction of $ X_{k} $; we thus identify $ \mathbb{R}^{d} $ with its Lie algebra in the usual way. For convenience, denote $ \alpha_{X_{k}} $ simply by $ \partial_{k} $. Next, for any multi-index $ \mu = (\mu_{1},\ldots,\mu_{d}) \in \mathbb{N}_{0}^{d} $ , let $ \partial^{\mu} $ denote the higher-order partial derivative $ \partial_{1}^{\mu_{1}} \cdots \partial_{d}^{\mu_{d}} $. Then equip $ A^{\infty} $ with the seminorms $$ \forall (j,k) \in \mathbb{N} \times \lbrace 1,\ldots,d \rbrace ~~ \& ~~ \forall a \in A^{\infty}: \quad \| a \|_{j,k} \stackrel{\text{def}}{=} \sup_{1 \leq i \leq j} \sum_{|\mu| \leq k} \frac{\| \partial^{\mu} a \|_{i}}{\mu!}, $$ where $ |\mu| \stackrel{\text{def}}{=} \mu_{1} + \cdots + \mu_{d} $ and $ \mu! \stackrel{\text{def}}{=} \mu_{1}! \cdots \mu_{d}! $.

My question is: As we are applying partial derivatives to $ a \in A^{\infty} $, are we identifying $ a $ with the function $ f_{a}: \mathbb{R}^{d} \rightarrow A $ defined by $ {f_{a}}(\mathbf{x}) \stackrel{\text{def}}{=} \alpha(\mathbf{x},a) $?

Thank you very much in advance!

Answer by Leonard for A novice question on Quantum Mechanics

2012-12-24T07:12:48Z

Let $ |A \rangle $ and $ |B \rangle $ be two non-zero vectors of a Hilbert space $ \mathcal{H} $ that belong to two different one-dimensional subspaces of $ \mathcal{H} $. According to Dirac, $ |A \rangle $ and $ |B \rangle $ represent two different quantum states.

Now, consider two non-trivial superpositions of $ |A \rangle $ and $ |B \rangle $: $$ |R_{1} \rangle := a_{1} |A \rangle + b_{1} |B \rangle \quad \& \quad |R_{2} \rangle := a_{2} |A \rangle + b_{2} |B \rangle, $$ where non-trivial means that $ (a_{1},b_{1}),(a_{2},b_{2}) \in \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. If $ |R_{1} \rangle $ and $ |R_{2} \rangle $ are to represent the same quantum state, then they must lie in the same one-dimensional subspace of $ \mathcal{H} $, i.e., they must be non-zero scalar multiples of each other. Knowing this, write $ |R_{2} \rangle = \lambda |R_{1} \rangle $, where $ \lambda \in \mathbb{C}^{\times} $. As $ \lbrace |A \rangle,|B \rangle \rbrace $ is a linearly independent subset of $ \mathcal{H} $, it follows that the condition $ (a_{2},b_{2}) = \lambda (a_{1},b_{1}) $ must be met. This condition does not hold for all choices of $ (a_{1},b_{1}) $ and $ (a_{2},b_{2}) $ in $ \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. Therefore, in general, $ |R_{1} \rangle $ does not represent the same quantum state as $ |R_{2} \rangle $.

On similar concepts in mathematics whose similarity is a non-trivial fact.

2012-12-17T21:43:26Z

Recently, while undertaking a study of commutative algebra, I learned three concepts: (i) a local ring, (ii) a regular local ring and (iii) a regular ring.

At the end, I found myself asking this seemingly naïve question: Are regular local rings the same objects as local rings that are regular? At first, I thought, "My mind must be acting stupid again." However, upon further analysis, it turned out that the answer to my question was non-trivial after all.

One direction, namely proving that a local ring that is regular is actually a regular local ring, is not very hard to establish. Indeed, it can be assigned as a homework problem in an undergraduate abstract algebra course. The key observation is that for a local ring $ (R,{\frak{m}}) $, upon localization at $ {\frak{m}} $, we obtain $ R_{\frak{m}} = R $. This is because $ R \setminus {\frak{m}} $ is precisely the set of units of $ R $. Hence, by the definition of regular ring, we see that $ (R,{\frak{m}}) $ is a regular local ring.

The other direction is a well-known (in my opinion, highly) non-trivial result in homological algebra, which states that the localization of a regular local ring at any prime ideal is still a regular local ring. By the definition of regular ring once again, regular local rings are therefore local rings that are regular.

I am wondering, are there any pairs of concepts in other areas of mathematics that look so similar that their similarity may be mistaken for tautology but, in reality, can only be established by a hard proof?

Answer by Leonard for Statements which were given as axioms, which later turned out to be false.

2012-12-19T10:52:47Z

In the mathematical theory of social welfare, it is possible to create a list of axioms that lead to a contradiction. For example, in voting theory, the following axioms for a voting system are considered reasonable in order for the system to qualify as being fair:

Each voter can have any set of rational preferences. This requirement is called “universal admissibility”.
If a voter prefers Candidate A to Candidate B, and Candidate B to Candidate C, then he/she prefers A to C. This requirement is called “transitivity”.
If every voter prefers A to B, then the group prefers A to B. This is sometimes called the “unanimity” condition.
If every voter prefers A to B, then any change in preferences that does not affect this relationship must not affect the group preference for A over B. For example, if a set of historians unanimously decides that Abraham Lincoln was a better president than Chester A. Arthur, a changing opinion of Bill Clinton should not affect this decision. This more subtle requirement is called “independence from irrelevant alternatives”.
There are no dictators. In other words, no voter exists whose preferences determine the preferences of the whole group.

The mathematical economist Kenneth Arrow showed in a landmark paper (stemming from his PhD thesis) that one obtains a contradiction if all five assumptions are assumed to hold. In fact, Assumptions (1) - (4) imply the existence of a dictator. However, these assumptions seem fairly reasonable and consistent, so the fact that they are contradictory is why Arrow named his paper “A Difficulty in the Concept of Social Welfare”. His result is known nowadays as Arrow's Impossibility Theorem.

Answer by Leonard for Statements which were given as axioms, which later turned out to be false.

2012-12-19T05:33:04Z

Another example from real analysis would be the question of the pointwise convergence of the Fourier series of a continuous function (defined on a closed interval). Many people, including Dirichlet and even the master rigorist Weierstrass himself, believed that the Fourier series of such a function converges pointwise everywhere to the function itself. Some clung on to this belief so strongly that they even viewed it as an infallible axiom.

Hence, one can imagine the great upset when, in 1876, Paul du Bois-Reymond proved the existence of a continuous function whose Fourier series diverges at a point. His proof is non-constructive and uses a method called the principle of condensation of singularities. I have absolutely no idea how the method works, but I do know of a very common proof that uses the Baire Category Theorem (using the Baire Category Theorem, one can also prove the existence of continuous functions that are not differentiable at any point).

After the dust had settled in the wake of du Bois-Reymond's seismic discovery, people started fervently believing that there should exist a continuous function whose Fourier series diverges everywhere - an opinion that lay on the other extreme! Andrei Kolmogorov inadvertently lent support to this claim by exhibiting, in 1926, an $ {L^{1}}([- \pi,\pi]) $-function whose Fourier series diverges everywhere. However, there was great upheaval once more in Fourier-land when the combined efforts of Lennart Carleson and Richard Hunt in the late 1960's showed that the Fourier series of any $ f \in {L^{p}}([- \pi,\pi]) $ converges almost everywhere to $ f $, for all $ p > 1 $ (this result subsumes the case of continuous functions). During an interview with the AMS, Carleson revealed that he had originally tried to disprove his result (pertaining to $ p = 2 $), but in the end, his failure to produce a counterexample convinced him that he should be working in the other direction instead.

Therefore, in the field of Fourier analysis, viewpoints have changed and cherished beliefs have been destroyed - twice.

Spectra of elements of a Banach algebra and the role played by the Hahn-Banach Theorem.

2012-12-19T02:50:06Z

This problem was posed on Math StackExchange some time ago, but it did not garner any solutions there. I think that it is interesting enough to be posed here on Math Overflow, so here it goes.

Let $ \mathcal{A} $ be a unital Banach algebra over $ \mathbb{C} $, with $ \mathbf{1}_{\mathcal{A}} $ denoting the identity of $ \mathcal{A} $. For each $ a \in \mathcal{A} $, define the spectrum of $ a $ to be the following subset of $ \mathbb{C} $:

$$ {\sigma_{\mathcal{A}}}(a) \stackrel{\text{def}}{=} \lbrace \lambda \in \mathbb{C} ~|~ \text{$ a - \lambda \cdot \mathbf{1}_{\mathcal{A}} $ is not invertible} \rbrace. $$

With the aid of the Hahn-Banach Theorem and Liouville's Theorem from complex analysis, one can prove the well-known result that $ {\sigma_{\mathcal{A}}}(a) \neq \varnothing $ for every $ a \in \mathcal{A} $. All proofs that I have seen of this result use the Hahn-Banach Theorem in one way or another (a typical proof may be found in Walter Rudin's Real and Complex Analysis). Hence, a natural question to ask would be: Can we remove the dependence of this result on the Hahn-Banach Theorem? Is it a consequence of ZF only? Otherwise, if it is equivalent to some weak variant of the Axiom of Choice (possibly weaker than the Hahn-Banach Theorem itself), has anyone managed to construct a model of ZF containing a Banach algebra that has an element with empty spectrum?

A Question About Free Resolutions

2012-12-11T02:20:09Z

I would warmly appreciate it if someone could tell me whether the following question has an affirmative answer. I am new to the field of commutative algebra, so I am simply trying to fill in some (huge) gaps. Thanks!

Let $ (R,{\frak{m}}) $ be a Noetherian local (commutative unital) ring. Let $ I $ be an ideal of $ R $ with minimal generating set $ \lbrace x_{1},\ldots,x_{n} \rbrace $, and let $ \beta: R^{n} \rightarrow I $ be the surjective $ R $-linear map defined by $ \beta(r_{1},\ldots,r_{n}) = r_{1} x_{1} + \cdots + r_{n} x_{n} $. Viewing $ I $ as an $ R $-module, does there exist a free resolution of $ I $ of the form $$ 0 \longrightarrow R^{n-1} \stackrel{\alpha}{\longrightarrow} R^{n} \stackrel{\beta}{\longrightarrow} I \longrightarrow 0, $$ where the map $ \alpha $ is left-multiplication by some matrix $ M \in {\text{M}_{n \times (n-1)}}(R) $?

Answer by Leonard for A certain theorem about finite-dimensional Lie algebras over an algebraically closed field with zero characteristic.

2012-12-06T22:59:43Z

Update

The bug in the example in Salvo's answer above has finally been fixed! It is now entirely correct.

My original concern was that, due to the requirement that the Jacobi Identity be satisfied, an infinite-dimensional Lie algebra (or even a finite-dimensional one) cannot be defined simply by playing with relations among basis vectors (which is what Salvo and I did initially, thus leading to a number of false starts). Hence, one needs to either work with "natural" examples of infinite-dimensional Lie algebras (as Professor Humphreys has mentioned in his comment below the wording of my question) or to struggle really hard with the Jacobi Identity in order to generate "non-natural" examples.

A certain theorem about finite-dimensional Lie algebras over an algebraically closed field with zero characteristic.

2012-12-05T06:42:47Z

Using Engel's Theorem and Lie's Theorem, one can easily establish the following result:

Let $ \frak{g} $ be a finite-dimensional Lie algebra over an algebraically closed field $ \mathbb{F} $ of characteristic $ 0 $. If $ \frak{g} $ is solvable, then $ [{\frak{g}},{\frak{g}}] $ is nilpotent.

In order to apply the two theorems stated at the beginning, one must assume that (i) $ \mathbb{F} $ is algebraically closed, (ii) $ \mathbb{F} $ has characteristic $ 0 $, and (iii) $ \frak{g} $ is finite-dimensional. If we relax each of these three conditions in turn, are there certain well-known counterexamples?

The Speed of Gravitational Waves in General Relativity

2012-11-10T21:37:44Z

Is it possible to mathematically prove that the speed of gravitational waves in general relativity equals the speed of light, without linearizing the Einstein Field Equations? The approach via the linearization of the EFE's, which is used in many books on relativity, does not seem to provide an exact proof that disturbances in spacetime propagate at the speed $ c $.

Answer by Leonard for Request: intermediate-level proof: every 2-homology class of a 4-manifold is generated by a surface.

2012-08-30T08:47:39Z

I do not have enough points to post a comment, so I will have to pose my question here. Could someone explain why the pre-image of $ \mathbb{CP}^{1} $ should represent the class? How does one ensure that the pushforward (with respect to the inclusion map into the original 4-manifold) of the top-homology class of the pre-image is equal to the given 2nd-homology class? It would be great if someone could show this in full detail.

Comment by Leonard

2012-12-24T21:32:53Z

Hence, you are right. As Professor Andreas Blass has mentioned in his comment below the wording of your question, superposition is a binary operation on non-zero vectors, not on states, which are equivalence classes of non-zero vectors.

Comment by Leonard

2012-12-24T21:22:50Z

Hi Ryan. We always have $ a_{1} |A \rangle \sim a_{2} |A \rangle $ and $ b_{1} |B \rangle \sim b_{2} |B \rangle $, yet we may still end up with $ |R_{1} \rangle \nsim |R_{2} \rangle $. Superposition of vectors is not a quantum-state-preserving binary operation on $ \mathcal{H} \setminus \lbrace 0_{\mathcal{H}} \rbrace $, unless $ |A \rangle \sim |B \rangle $. You can prepare a myriad of quantum states from just two distinct quantum states. This principle is important in quantum computing, in which one can produce infinitely many states from a fixed basis of a finite-dimensional Hilbert space.

Comment by Leonard

2012-12-19T11:21:41Z

Thank you very much for the references!

Comment by Leonard

2012-12-19T06:25:55Z

@Alexandre: I will definitely follow your last piece of advice. It seems to be the most practical.

Comment by Leonard

2012-12-19T05:03:20Z

I wasn't referring to Qiaochu's question, but it sure is a surprise to see that his asks almost the same thing.

Comment by Leonard

2012-12-19T04:57:51Z

@Alexandre: I have the same problem all the time. Do you have any tips for making MathJax process syntax correctly? Sometimes, MathJax cannot read what I've written, although everything is correct.

Comment by Leonard

2012-12-19T02:16:30Z

Did Jean-Pierre Serre's GAGA not change the way that complex analysts and algebraic geometers look at each other's field?

Comment by Leonard

2012-12-19T02:13:33Z

Very interesting indeed! Looks trivial, but after looking at the actual proof, I am deeply impressed by its depth.

Comment by Leonard

2012-12-11T07:03:59Z

@Sasha: Your comment is intriguing. Could you kindly direct me to some references where this identity is proven?

Comment by Leonard

2012-12-11T02:40:56Z

I see. Then may I know under what conditions on $ I $ will such a free resolution exist?

Comment by Leonard

2012-12-08T08:44:08Z

Would you be able to provide more references for the fact that pullback in homology corresponds to pullback in geometry? Thanks!

Comment by Leonard

2012-12-07T22:44:33Z

@Salvo: I am unsure of how the semidirect product of a Lie algebra by a left module is defined, but I do know how to form the semidirect product of two Lie algebras. Please correct me if I am wrong to say the following. In your example, you are treating $ \mathbb{F}[t] $ as a Lie algebra with the trivial Lie bracket, and you are defining a Lie-algebra homomorphism $ \varphi: H \rightarrow \text{Der}(\mathbb{F}[t]) $ using the actions of $ x $, $ y $ and $ z $ that you have described (as $ \mathbb{F}[t] $ has a trivial Lie bracket, these actions are automatically Lie-algebra derivations).

Comment by Leonard

2012-12-07T19:41:05Z

The example is excellent. Thanks! It turns out that things are not as easy as they seemed at first sight. :)

Comment by Leonard

2012-12-06T22:08:18Z

Thank you, Professor Humphreys!

Comment by Leonard

2012-12-05T16:00:32Z

Yes, I just realized that one can remove the 'algebraically closed' condition. Thank you for the comment anyway!