The n-Category Café

As far as I can tell, the amount of ejected material, tidal debris, etc, will depend on the collision scenario (orbital energy of the galaxies) as well as on their mass distribution (baryonic and non-baryonic) and possibly on other finer details such as the amount of diffuse mass of the Local Group (these latter are important for taking into account dynamical friction effects).

I have never simulated such a detailed collision between the Milky Way and Andromeda, but Cox and Loeb did.

(My previous work with N-body simulations focused on merging elliptical-like objects in the context of the origin of the “fundamental plane” relation, and were intended as to study the overall features of the mergers, not their detailed interactions. If memory serves, however, I could have about ~ 15% of mass ejections in some cases, or even more, depending on the “violence” of the merger, that is, how strongly the gravitational potential would fluctuate during the collision).

Christine

“Thanks! And this is close to one of my naive questions: “why everything is abelian?”. I mean that the large part of mathematics is built on the notion of abelian (additive) group: rings, fields, bodies, modules, vector spaces, etc.”

I think the answer is: for historical reasons only.

There are mathematical fields (like analysis in metric spaces, see the work of Gromov) where the natural notion replacing a vector space is a simply connected Lie group whose Lie algebra admits a positive graduation (aka Carnot group). A particular example is the Heisenberg group.
See http://xxx.arxiv.org/abs/0705.1440
for a definition of linearity in this context.

200 years ago there was a single geometry:
euclidean. Is it risky to say that in 100
years from now we shall have non-euclidean analysis, which is not based on abelian groups?

Osman said, “why everything is abelian?”

Some people have argued that any upper bounds on the amount of non-abelianness around us should be pretty weak. But let me try to actually give some upper bounds.

The basic idea is that objects which are non-abelian often fail to admit higher operations. For example, abelian groups can admit ring structures, but a bilinear map on a general group factors through its abelianization.

Bergman and Hausknecht’s book “Cogroups and co-rings in categories of associative rings” is a pretty extensive reference for things like this. Part of it is based on the work of Tall–Wraith Let me try to say a few words about what they do, as far as I understand it.

In what sense is a (possibly non-commutative) ring naturally the kind of object that knows how to act on an abelian group? The answer is that is represents a comonad on abelian groups. More precisely, we say an endofunctor on the category of abelian groups is representable if its composition with the forgetful functor to sets is representable. If $R$ is an abelian group, then the functor $\mathrm{Hom}(R,-)$ naturally takes values in the category of abelian groups, and a ring structure on $R$ is the same thing as a comonad structure on the functor.

The interesting thing is that Bergman and Hausknecht show that it’s hard to do similar things in non-abelian situations. For example, there is a theorem of Kan that says that the category of representable endofunctors on the category of groups is equivalent to the category of pointed sets. It follows that the category of representable comonads on the category of groups is the same thing as a monoid in the category of pointed sets, which is the same thing as a monoid. In other words, the only thing that naturally knows how to act on groups is just a monoid. Whereas in the category of abelian groups, we went up to the next level and got rings.

A similar thing happens at the next level. What are the representable comonads on the category of rings? What about the category of commutative rings? Here things are a little more interesting, because there are objects that know how to act on rings, but not by ring endomorphisms! For instance, a Lie algebra knows how to act on a commutative ring. Such a thing is just a homomorphism from the given Lie algebra to the Lie algebra of derivations of the ring. In fact, more generally, any cocommutative bialgebra knows how to act on a commutative ring. In the case of Lie algebras, you get the commutative ring representing the functor by taking the symmetric algebra of the enveloping algebra. In the case of a group, you take the symmetric algebra of its group algebra.

The great thing is that in the case of commutative rings, there are even more examples, ones that don’t come from bialgebras. For example, in positive characteristic, the Frobenius map knows how to act on a commutative ring. There is a representable comonad $T$ on the category of commutative $Z/\mathrm{pZ}$-algebras such that an action of $T$ on $R$ is nothing more than saying that the Frobenius map on $R$ is bijective. We could make another comonad that requires that the Frobenius map be the identity. Even more interesting, over $Z$, we could ask for lifts of the Frobenius map from characteristic $p$ to characteristic $0$.

But all these wonderful things can’t happen with noncommutative rings. Of course there is no non-commutative Frobenius map, at least in the straightforward sense. In fact, Bergman and Hausknecht prove that, at least when $F$ is $Q$ or $Z/\mathrm{pZ}$, the only representable comonads on the category of $F$-algebras are ones coming from bialgebras combined with ones coming from anti-endomorphisms, ie maps from an algebra to its opposite. (Admittedly, they don’t address the situation over $Z$, which is much more important.)

Now that I’ve said all that, they also give some evidence that goes against this point of view. For example, even though the only representable comonads on the category of groups come from boring monoids, there are actually representable comonads on the category of monoids that don’t come from monoids. For example, it’s possible to endow a monoid $M$ with a “right-inverse” operator. This is an anti-endomorphism $r:M\to M$ satisfying the identity $m\cdot r(m)=1$ and such that $r$ composed with itself is the identity. You can then use this structure to make other representable comonads. In fact, they give a complete description of all the representable comonads, and they have a similar flavor involving left and right inverse operators. It would be pretty interesting to study the next level of this hierarchy.

Categorifying negative numbers is actually tricky. John Baez talked about negative sets here, and gave a number of good references.

Another thing that people have tried is to use superalgebra; for example, instead of considering linear combinatorial species $F_0$ (i.e., species valued in vector spaces), consider “virtual” species valued in ${\mathbb{Z}}_2$-graded vector spaces. A virtual species is given by a pair of linear species

$(F_0,F_1)$

which is to be thought of as a formal difference $F_0-F_1$ (hence, at the decategorified level, this involves passage to a Grothendieck group). In good cases, a virtual species may also come equipped with differential graded structure, and one may consider the derived category of DG-species as another method for dealing with categorified negatives.

But none of these methods is perfect.

Expressing the concept of “positive energy representation” in the language of groupoids sounds tricky, since it’s rather closely wedded to the complex numbers. For example, there’s no obvious notion of a positive-energy real representation of a Lie group.

To see this, consider the primordial example: the real line! A strongly continuous unitary (complex) representation $U$ of $ℝ$ has a skew-adjoint generator

$$A=\frac{d}{dt}U(t){\mid }_{t=0}$$

which then has

$$U(t)=\exp (tA).$$

The spectrum of $A$ lies on the imaginary line, so it makes no sense to say it’s positive.

Using the magic of $i$, we can then define a self-adjoint generator

$$H=iA$$

which has

$$U(t)=\exp (-itH)$$

Since the spectrum of $H$ lies on the real axis, it makes sense to declare $H$ positive when its spectrum lies in $[0,+\mathrm{\infty })$. We then say $U$ is a positive-energy representation.

Note how the ghost of Galois hovers over the proceedings, snickering quietly. If we’d picked $-i$ instead of $i$ as our square root of $-1$, what used to be a ‘positive-energy representation’ would now be declared ‘negative-energy’, and vice versa!

If we were working over the real numbers, we wouldn’t have positive-energy representations.

See how everything changes:

A strongly continuous orthogonal (real) representation $U$ of $ℝ$ has a skew-adjoint generator

$$A=\frac{d}{dt}U(t){\mid }_{t=0}$$

which then has

$$U(t)=\exp (tA).$$

The spectrum of $A$ lies on the imaginary line, so it makes no sense to say it’s positive. In fact, if any number $a$ lies in the spectrum, so must its complex conjugate $\overline{a}$!

So far, groupoidification seems to work most easily for the most ‘field-independent’ aspects of linear algebra. The concept of positive energy is very special to the complex numbers. So, if we want to groupoidify the theory of positive-energy representations, we have our work cut out for us.

So far, Jim Todd and I are much closer to groupoidifying the whole theory of unitary representations of finite-dimensional compact simple Lie groups — or equivalently, finite-dimensional holomomorphic representations of complex simple Lie groups. The concept of ‘positive energy representation’ doesn’t apply here.

In short: there’s a lot of work to do, and I’ve got to tell you about it before you can see where the frontier is… but ‘positive energy representations’ are a bit beyond the frontier, right now.

By the way, I mentioned this business about skew-adjoint versus self-adjoint generators in my discussion of real, complex and quaternionic quantum mechanics in week251:

Another special way in which $ℂ$ is better than $ℍ$ or $ℝ$ is that only for a complex Hilbert space is there a correspondence between continuous 1-parameter groups of unitary operators and self-adjoint operators. We always get a skew-adjoint operator, but only in the complex case can we convert this into a self-adjoint operator by dividing by $i$.

In the case of $ℝ$ there aren’t enough square roots of -1. In the case of the quaternions, $ℍ$, there turn out to be ‘too many’ — and the problem is, they don’t commute. So, if $A$ is linear, $H=iA$ is not!