3
$\begingroup$

Lingua Franca links relativity and quantum theories with spectral geometry

Could someone give me a short synopsis of this article in laymens terms? What implications does this have in the physics community? Is this work ground breaking or just the start of something that might be?

From what I can understand this physicist related two types of maths that we use to model the world around us. Which has been hard to do, because the types of maths are incompatible with each other.

But what is this spectral geometry? How does it relate the two types of maths in relativity and quantum theory? What might those maths be? What about the two has made it hard to unite them?

Improve this question
$\endgroup$
5
  • $\begingroup$ It's just a press release, not a world changing discovery. Did you try googling 'spectral geometry'? $\endgroup$
    – user1504
    Commented Apr 30, 2013 at 12:19
  • $\begingroup$ Yeah, I'm afraid I can't understand it really. If this is a bad question just let me know so I can delete it. I didn't think it really made sense. I just wanted to know what this physicist did in laymens terms, if possible to do so. $\endgroup$
    – KDecker
    Commented Apr 30, 2013 at 12:26
  • $\begingroup$ I don't think it's a bad question (or rather, I think some of the questions are good, and some just result from taking a press release too seriously), but you're asking for a lot of exposition, so I wanted to know that you'd put in some effort yourself. $\endgroup$
    – user1504
    Commented Apr 30, 2013 at 15:31
  • $\begingroup$ What I usually do is read every article on phys.org I can, I then search for hours on each article trying to understand it fully and get some more sources. This one has eluded me, I just like to get a good idea of what the article is saying.--- I didn't mean it to sound like I think the article is prolific, but from what I know it has been a long standing challenge to unite relativity and mechanics together for the Grand Unified Theory of Everything. --- I don't understand the maths or even the maths concepts to try and search them myself $\endgroup$
    – KDecker
    Commented Apr 30, 2013 at 15:40
  • $\begingroup$ I just kinda wanted something along the lines of "this lady showed via spectral geometry that a non-continuous manifold can be represented by a continuous manifold math" (I have no idea what I am saying, just pulled words from the article) But with something like that I can search each idea and try to relate them, figure something out, then absolutely destroy the idea when I try to tell my friends. $\endgroup$
    – KDecker
    Commented Apr 30, 2013 at 15:42

2 Answers 2

Reset to default
5
$\begingroup$

Spectral geometry is one of the many ways mathematicians think about geometry. The general idea is that if you have some manifold equipped with a metric, you can cook up some canonical differential operators. These operators can be thought of as linear operators, acting on (infinite-dimensional) vector spaces of functions, tensors, spinors, and the like. Each such linear operator will have a set of eigenvalues. Spectral geometry is concerned with relationships between these eigenvalues and the geometry of the manifold you started with

The most obvious linear operator to associate to a metric is the Laplacian, which is a linear operator on the space of functions on the manifold. In the early/middle part of the 20th century, mathematicians started wondering "If you know the set of eigenvalues of the Laplacian, can you reconstruct the manifold?", or as Mark Kac famously put it: "Can one hear the shape of a drum?"

The answer is no; the set of eigenvalues alone doesn't let you reconstruct the manifold and its metric. The map which sends a manifold with metric to the set of eigenvalues of the Laplacian is not invertible.

But it's such a pretty idea that people haven't given up on it. Alain Connes, for example, figured out that the answer to a slightly different question is "yes". If you have a commutative spectral triple (basically, the Dirac operator on a compact spin manifold), you can reconstruct the metric from this data.

The physicists interviewed in the linked article are trying a slightly different variation on the spectral geometry problem. They're considering systematic finite-dimensional approximations to the "derivative" of the map $F$ which sends a manifold to its set of eigenvalues of its Laplacian acting on tensors of low-degree, and trying to show that these approximations are invertible. This should let them show that this map $F$ is invertible in small regions of the space of manifolds.

It's a nice idea, and looks like some fun experimental mathematics. Trying to write down a theory of gravity in explicitly gauge invariant terms is also a good idea. But I'd be quite surprised if this line of thinking bears any fruit. It seems more likely to me that we need some essentially new physical ideas than a clever way of rewriting what we already have.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Ok this might come out a little jarbled. Ill start low. I've been trying to understand the Lapacian, which from calculus I think it in laymens terms means the derivative of a point on a function. I learned derivatives with just a line on a graph, so I figure the eigenvalues of the lapacian of the domains of the functions that represent the physical shape of the drumhead, means.. The lapacian(derivative) of a point on the drumhead makes sense to me(the inclination/slope of that point in space), put all these in a matrix, get the eigen value for that drumhead shape, one of the many it can be. $\endgroup$
    – KDecker
    Commented Apr 30, 2013 at 22:01
  • $\begingroup$ (Con't)Given all these eigenvalues for each shape the drumhead can be in, can we find the size of the manifold it represents? In the drum case, a circle? I did take a matrix math class, and if I remember the eigenvalue is unique to the matrix it is found from? So given all unique eigen values for each shape of a manifold, can another manifold have the same set of eigenvalues or not? And No as you said? $\endgroup$
    – KDecker
    Commented Apr 30, 2013 at 22:02
  • $\begingroup$ I'm afraid that you've gone astray. The concept I'm describing is roughly one layer of abstraction up. There's a linear map $L$, which sends a function $f$ to the function $Lf$ which is defined by $Lf(x) = \sum_{i,j} g^{ij}(x) \partial_i \partial_j f(x)$. (I'm afraid you were astray already by the phrase "derivative of a point of a function"; I'm not sure that means anything.) The eigenvalues in questions are eigenvalues of this operator $L$. $\endgroup$
    – user1504
    Commented Apr 30, 2013 at 22:54
  • $\begingroup$ Yeah. Well thank you first off. And last question, seeing the terms I phrased the second question in; could you say any thing short enough to fit in a comment something that sums any of this up? Otherwise, I shall never touch physicsStack again unless I know what I am saying. $\endgroup$
    – KDecker
    Commented May 1, 2013 at 0:51
  • $\begingroup$ well, the answer to the original "can you hear the shape of a drum" question is "No, different manifolds can have Laplacians with the same set of eigenvalues". Such manifolds are called 'isospectral'. $\endgroup$
    – user1504
    Commented May 1, 2013 at 1:45
-1
$\begingroup$

What I will comment here is supplementary to user1504's answer and what is not clear about iso-spectrality (iso for same and spectral for eîgenvalues) and spectral triple.

I have been working on a different topic but astonishingly found connections between spectral geometry and it. I have been working on Non-Hermitian Quantum Mechanics and a recent result suggests that Non-Hermitian operators having Real Eigenvalues 'always' have a sort of brother, a Hermitian operator with same set of eigenvalues ( these kind of operators are called iso-spectral)! A somewhat deeper thought suggests correctly that, they should be related by Similarity Transformation. Moreover, their state spaces form Hilbert space (A complete orthogonal vector space) with different metrics though (a metric is something similar to a length scale).

You can check all that User1504 suggested by taking a complex 2*2 matrix instead of using a differential equations and doing hardcore math. Just take a 2*2 general matrix make it Hermitian, perform a Non-Unitary Similarity transform on it and you will see that the new matix has same eigenvalues ( i.e. it is isospectral) but it has now become Non-Hermitian ! ( Why does this happen ?, well because similarity transformation always preserve eigenvalues, a small google search will give a lot of links to it's proof) This directly suggests what Alain Connes tries to say about spectral triple, A complete information is embedded not only in spectrum but your measurement scale that is Metric which is usually ignored and the algebra you are using for example C* in my case.

You can refer to Appendix of my paper if you want. Note just skip to the Appendix and everything will be clear. http://vixra.org/abs/1403.0047

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.