Everyone knows what a group is. A group \(G\) is a set of elements – operations – that may be composed ("multiplied" – most typically, we imagine products of matrices; the word "addition" and the sign \({+}\) is only appropriate for commuting, Abelian groups) so that the operation is associative,\[

(ab)c = a(bc),

\] that includes the unit element \(1\) obeying \(1a=a1=a\) for each \(a\in G\) and the inverse for each element.

Groups may be finite – a finite number of elements such as \(\ZZ_n\); infinite; countable, continuous, infinite-dimensional, and so on. For continuous groups, it's useful to consider the Lie algebras which may be imagined as the tangent space of the group manifold placed e.g. at the unit element, with the "commutator" operation that remembers how the group elements near the identity refuse to commute with each other.

More fun starts when we consider representations. I will talk about linear representations only. A linear or vector representation of the group \(G\) is a linear (vector) space \(V\) such that we know how to define the action of the group elements \(g\) on the vectors in \(V\):\[

g: \, v\mapsto g(v).

\] I could call the map \(\phi_g\) instead of \(g\) because the map \(V\to V\) is strictly speaking a different object than the "abstract" element of the group but physicists generally despise such redundant notation.

The maps \(g\) acting on \(V\) have to respect the structure of the group, i.e.\[

(g_1 g_2)(v) = g_1(g_2(v)).

\] If you transform \(v\) by the element \(g_2\) and then by \(g_1\), you get the same thing as if you transform it by \(g_1\cdot g_2\) where the dot represents the multiplication inside the group rather than a composition of actions on a particular vector space.

Moreover, it is a linear representation because the maps have to preserve the linearity of the vector space,\[

g(v_1+v_2) = g(v_1) + g(v_2).

\] If we pick a basis in \(V\), we may express the action of the linear operators \(g\in G\) by matrices. Now, the representations may be linear over complex numbers \(\CC\), real numbers \(\RR\), quaternions \(\HHH\), or something else. Pure mathematicians would love to consider various discrete number systems, too.

The complex numbers are more fundamental in representation theory because the real and pseudoreal representations may be presented as "special types of complex representations", i.e. as complex representations with some extra structure that is compatible with the structure (maps and axioms) we have already described.

Even if you don't understand some words in the previous paragraphs, try to follow this definition. A real or pseudoreal representation \(V\) is a representation of \(G\) such that it comes equipped with an additional antilinear map \(j\), the structure map, that obeys the antilinearity (the "anti-" adds the complex conjugation on the line below)\[

j(\lambda v) = \lambda^* j(v), \quad j(v_1+v_2) = j(v_1) + j(v_2)

\] and that obeys the condition that makes it a structure map\[

{\rm real}: j(j(v)) = +v, \,\,{\rm pseudoreal}: j(j(v)) = -v.

\] We sometimes summarize these conditions by writing \(j^2=\pm 1\) where the squaring means the composition of two maps (the action of the map on the result of the same map acting on a vector) and where we omit the vector \(v\) because \(j\) is also a sort of an "operator". I forgot that the structure map has to respect the group action i.e. obey\[

\forall g,v:\,\,j(g(v)) = g(j(v))

\] What does the existence of the structure map do for us?

Let's start with the real structure map, \(j^2=+1\). With this extra antilinear map \(j\), we may require the vectors \(v\in V\) to be invariant under \(j\) i.e. \(j(v)=v\). We know that the eigenvalues of \(j\) are \(\pm 1\) because \(j^2=\pm 1\). Note that these are eigenvalues with eigenvectors in a real vector space; we can't treat \(V\) as a complex (or more complicated) vector space from the viewpoint of \(j\) because \(j\) isn't a linear operator over \(\CC\); it's antilinear. But the complex conjugation does nothing for coefficients \(\lambda\in\RR\).

Fine. So the vectors obeying \(j(v)=+v\) may be called "real vectors" and the vectors obeying \(j(v)=-v\) are "pure imaginary vectors". We may choose a real basis of the real vectors of with respect to \(j\). Because \(jg=gj\) as we wrote above ("the structure map respects the group structure"), we may easily see that the matrices of \(g\in G\) relatively to this real basis will be real. After all, the map \(j\) may be interpreted as the complex conjugation of the coordinates written relatively to the real basis and the condition \(j(g(v))=g(j(v))\) means that it doesn't matter whether you first complex conjugate the vector and then act with the matrix; or you first act with the matrix and then complex conjugate the result. This identity implies that the matrix itself is real, of course.

So linear representations with the \(j^2=+1\) are "real representations" which means that in appropriate bases – the real bases – the matrices representing all \(g\in G\) are real matrices with entries in \(\RR\). Just to be sure, we may still choose "non-real bases" and it's often very useful – for example, spherical harmonics \(Y_{\ell m}\) for \(\ell=1\) and \(m\in\{-1,0,+1\}\) team up to give us a non-real basis of the 3-dimensional space of vectors with a natural action of the \(SO(3)\) group (and the \(SU(2)\) group, too) whose generators are, however, expressed by matrices that are not real (and they're not uniformly pure imaginary, either).

So the \(j^2=+1\) structure map tells us a simple thing: the matrices may be made real. What about the \(j^2=-1\) structure map? It's sort of cute that in this form, the structure map differs from the previous one "just by a sign". But what does it do? How can it be interpreted?

This \(j\) map can't be interpreted as a multiplication by \(i\) because it wouldn't be antilinear. It can't be interpreted as the multiplication by \(i\) composed with the complex conjugation, either, because it would obey \(j^2=+1\). For example, \(1\to i\to -i\to 1\to 1\).

However, this \(j\) map may be interpreted as the multiplication by the quaternion that is conveniently denoted by the letter \(j\), too! And I mean the multiplication from the right side. Effectively, this structure map tells us that we know not only how to multiply \(v\in V\) by complex scalar coefficients (because it's a vector space over \(\CC\)); we also know how to multiply them by \(j\), and because \(ij=k\) and because we know the previous two things, we also know how to multiply vectors by \(k\). Consequently, all the elements in \(v\in V\) may be interpreted as collections of quaternion-valued coordinates.

Just to be sure, the identity \(j^2=-1\) we need for the structure map holds simply because \(j^2=-1\) holds for the imaginary unit quaternion \(j\). And \(j\) commutes with the action of the group elements because we may imagine the group elements as acting from the left on \(v\) while the multiplication by \(j\) acts from the right – but the product is still associative (quaternions and matrices constructed out of them are associative). Finally, the map is antilinear and not linear because \(i,j\) anticommute:\[

(e+if) j = j(e-if) = j(e+if)^*.

\] That was the background. I want to dedicate the rest of this article to describe which representations of compact Lie groups – and the pseudoorthogonal groups – are complex, real, and pseudoreal.

These are the Dynkin diagrams for the simple compact Lie groups. They come in four infinite families A,B,C,D and five exceptional Lie groups written as E,F,G where the subscripts denote the "rank" – the maximum number \(k\) such that \(U(1)^k\) is a (commuting) subgroup, the so-called maximal torus (the Lie algebra of it is the Cartan subalgebra).

Which of the groups have complex, real, and pseudoreal representations? All the representations have real representations. For example, the adjoint representation – which is, as a linear space, the Lie algebra of the group itself – is always real. The same is true for the 1-dimensional "singlet" representation that doesn't transform at all. A funny fact is that complex representation – by which I mean representations that don't admit any \(j^2=\pm 1\) structure maps – occur exactly when the Dynkin diagrams have some symmetry or symmetries, when a nontrivial permutation of the nodes is able to give the same Dynkin diagram back.

Why? Because such a permutation of the nodes may be extended to an "outer automorphism" of the group, a relabeling that preserves the group product relationships (that's what an "automorphism" means) that can't be obtained by a conjugation \(g\to hgh^{-1}\) (that's when we call it "outer": the automorphisms given by simple conjugations are "inner"). But such an outer automorphism also has to act on the representations but because it's outer, it's actually exchanging irreducible representations with other irreducible representations (of the same dimension). That has to be possible if the representation is complex – because the complex conjugation produces a different (unitarily inequivalent) representation.

So let's look at all the Dynkin diagrams.

\(A_n=SU(n+1)\) has a left-right symmetry so it does have complex representations (inequivalent to their complex conjugate ones; equivalently, representations without a structure map). That's why groups \(SU(n+1)\) have fundamental representations that are \((n+1)\)-dimensional and complex, and their complex conjugate representations. We write them in bold face, e.g. \({\bf 3}\) and \(\overline{\bf 3}\) for \(SU(3)\), the group we often encounter in QCD. If we talk about the "color \(SU(3)\)", the two 3-dimensional representations are those in which the 3 colors of a quark and 3 anticolors of an antiquark transform.

Well, there is an exception. \(A_1=SU(2)\) only has one node and the left-right reflection of the Dynkin diagram is actually a trivial permutation (one that does nothing). So we haven't found a nontrivial outer automorphism and as a result, the fundamental representation \({\bf 2}\) of \(SU(2)\) is actually not complex: it is pseudoreal! It is equivalent to its complex conjugate, \({\bf 2} = \overline{\bf 2}\). That's why there is only 1 type of a spinor in 3 spatial dimensions.

In \(SU(n)\) groups with \(n\gt 2\), the fundamental representations are complex and we may build lots of other complex and real representations by taking tensor products. I haven't discussed it but there's a cute way to find a structure map on \(V\otimes W\), the tensor product of two representations. It may be written schematically as \(j=j_v\cdot j_w\) and the funny thing is that it's trivial to calculate \[

j^2 = j_v^2\cdot j_w^2.

\] So we just multiply the signs \(j^2\) from the two factor representations \(V,W\) to get \(j^2\) of their tensor product! It follows that the tensor product of two real reps is real; the tensor product of one real and one pseudoreal rep is pseudoreal; and the tensor product of two pseudoreal reps is real again because the two minus signs cancel. (Of course, the tensor product of a complex rep with anything is complex.)

For example, the tensor product \({\bf 2}\otimes {\bf 2}\) of \(SU(2)\) representations is a 4-dimensional representation – the dimensions are

*always*counting \(n\) when the representations are visualized as \(\CC^n\) because complex numbers are really fundamental everywhere here and they're the right way to count the dimensions (the real representations are allowed to be complexified i.e. doubled so the number of assumed complex coordinates is the same as the number of the real coordinates; the quaternionic representations' coordinates are split to pairs of complex numbers, so the declared dimension is twice the number of the quaternionic coordinates).

But the 2-dimensional representation has \(j^2=-1\) which means that a tensor product of two of them has \(j^2=+1\). And indeed, the tensor product splits to \({\bf 3}\oplus {\bf 1}\) which is a triplet (vector) and a singlet (scalar). Both of these irreps are real, just like we needed.

Let's continue. \(B_n=SO(2n+1)\). This is an orthogonal group. There's no symmetry of the Dynkin diagram; note that a symmetry of the Dynkin diagram must also preserve the orientation of the double and triple links. So the representations can't be complex. And indeed, all representations of \(SO(2n+1)\) may be constructed out of the fundamental \((2n+1)\)-dimensional vector representation which is real; or the spinors which are real or pseudoreal for an odd spacetime dimension, as discussed later.

The next Dynkin diagram describes the compact symplectic group \(C_n=USp(2n)\) which may also be written as \(U(n,\HHH)\), a unitary group over quaternions. (The Hermitian conjugation involves complex conjugation that changes the signs of all three imaginary units.) Again, the Dynkin diagram has no symmetry which means that there are no complex representations. The fundamental representation is actually pseudoreal, a vector with \(n\) quaternionic components.

When I said that \(B_n\) and \(C_n\) had no symmetries, I meant this claim for \(n\geq 2\) including the case \(n=2\). The left-right reflection of \(B_2\) is \(C_2\): they're the same Dynkin diagrams which means \(SO(5)=USp(4)\) but it changes nothing about the absence of complex representations.

Finally, \(D_n=SO(2n)\) has a symmetry – when the diagram forks into the two last steps, up and down, you may interchange up and down. That's why \(SO(2n)\) has complex representations. The \(2n\)-dimensional fundamental vector is clearly real, not complex, but the group has complex spinor representations, as discussed later. Well, it's actually only true for \(SO(4k+2)\), odd multiples of two. The \(SO(4k)\) group only has real or pseudoreal representations but they come in pairs of mutually inequivalent spinor representations of the same type ("chirality"). For \(SO(4k+2)\), these two chiral reps are complex conjugate to each other; for \(SO(4k)\), each of them is complex conjugate to itself because they're both real or they're both pseudoreal.

Finally, among the five exceptional Lie groups' Dynkin diagrams \(E_6,E_7,E_8,F_4,G_2\), only \(E_6\) has a symmetry – it's the left-right-symmetric tree with equally long branches on the left side and the right side. That's why \(E_6\) is the only compact exceptional Lie group that has complex representations. The basic ones are \({\bf 27}\) and \(\overline{\bf 27}\), the fundamental representation and its complex conjugate. All other exceptional groups only have real or pseudoreal representations – starting from the fundamental ones.

Every group has some real representations – the singlet and adjoint irreps are real, after all. The existence of pseudoreal irreps is somewhat more special. Among the compact simple Lie groups, they exist for \(SU(4k+2)\), \(Spin(8k+3)\), \(Spin(8k+5)\), \(USp(2n)\), \(Spin(8k+4)\), and \(E_7\). The pseudoreal representations of the \(Spin\) groups start with the pseudoreal spinors discussed later; the symplectic group has a quaternionic fundamental representation "by construction", \(E_7\) is the only exceptional group with pseudoreal representations because this \({\bf 56}\) is tensor-multiplied by another pseudoreal \({\bf 2}\) in the decomposition of \({\bf 248}\) of \(E_8\) under the maximal \(E_7\times SU(2)\) subgroup; and the \(SU(4k+2)\) pseudoreal irreps may be related to by doubling to the spinors of \(Spin(8k+4)\).

This text isn't primarily about spinors: see e.g. this older one. But for the sake of completeness, I want to discuss the reality or pseudoreality or complexity of the spinor representations.

First, in odd space or spacetime dimensions, there is only one type of a spinor. It's because the left-right reflection (changing a left hand to a right hand) may be represented by minus unit matrix: its determinant is \(-1\) if the size of the matrices is odd! So we have a particularly simple matrix representing "parity". This minus unit matrix clearly maps a vector in a representation to the vector in the same representation (minus the original vector). That's why the mirror image of a spinor representation is the same spinor representation: there's only one spinor type in odd dimensions and no inequivalent one.

In even space or spacetime dimensions, one may define the chirality operator most notoriously known as \(\Gamma_5\) in 4 dimensions. In this way, the Dirac spinor representation in \(2n\) dimensions is effectively promoted to a spinor that knows how to transform under the group \(SO(2n+1)\) in a higher dimension! However, the Dirac spinor isn't irreducible. We may see that \(\Gamma_5\) commutes with all the generators of the 4-dimensional rotations or Lorentz transformations \(\Gamma_{\mu\nu}\) which is why we may split the Dirac spinor representation to pieces with \(\Gamma_5=\pm 1\) which are the two allowed values because \((\Gamma_5)^2=1\) in some convention of \(i\) factors. These two halves of the Dirac representation are the chiral or Weyl representations. We may call one of them "left-handed spinor" and the other is a "right-handed spinor". The naming follows some messy conventions. The two chiral spinors are interchanged by a left-right flipping transformation. But in an even dimension, this parity cannot be represented by the minus unit matrix: the determinant of the latter is equal to \((-1)^{2n}=+1\).

But I still haven't said anything about the complexity of all these spinor representations. The spinor representations of \(SO(8k+m)\) are complex (c), real (r), or pseudoreal (q) according to this table – yes, the periodicity of the pattern is eight (various related facts are known as the Bott periodicity):\[

\begin{array}{|c|c|c|c|c|c|c|c|c|}

\hline

k&1&2&3&4&5&6&7&8\\

\hline

{\rm type}&r&c&q&q&q&c&r&r\\ \hline

\end{array}

\] If you start with \(Spin(2)\) and increase the argument, the letters say C-Q-Q-Q-C-R-R-R-C-Q-Q-Q-C-R-R-R, and so on. You should sing this song at least for 14 minutes to memorize it. ;-)

You might think that the type of the representation may only depend on the parity of \(n\) in \(Spin(n)\) – whether it's even or odd. But that's not the case. It also depends on whether it's a multiple of four or not. And in fact, what it is modulo 8. Note that the Cartan subalgebra that matters divides the spacetime dimension by two and to find the complexity or reality of the representation, you need to divide the Cartan generators to two groups again, read them backwards, and these extra operations in the proof – that I will omit – make the periodicity equal to a whopping eight.

However, if you believe that the periodicity is eight, it's not hard to verify that the types of the representations are given by the table above. Start with \(Spin(1)\). There's nothing to rotate so you may actually make the singlet representation real. This looks like too simple an example and you may want to work with \(Spin(9)\) instead. Its spinor is real because it's the direct sum of the two spinors of \(Spin(8)\) and we will see why they're real momentarily.

\(Spin(2)=U(1)\) is just the group of rotations by one angle. Because it's the same as \(U(1)\), it's clear that a single complex number, one that changes the phase, is a representation. So the fundamental spinor representation is complex.

I have already discussed \(Spin(3)\) because it's isomorphic to \(SU(2)\) and the fundamental 2-component spinor is pseudoreal – the only \(SU(m)\) Dynkin diagram that has no symmetries (outer automorphisms).

\(Spin(4)\) is another special case because it's isomorphic to \(SU(2)\times SU(2)\): the six generators of \(SO(4)\) may be split to self-dual and anti-self-dual ones. Each group forms its own \(SU(2)\) algebra. The spinors of \(Spin(4)\) are therefore either the 2-component representation of the first \(SU(2)\), the left-handed spinors, or the 2-component representation of the second \(SU(2)\), the right-handed spinor. Those are pseudoreal representations, as explained in the previous paragraph, but you should notice that \(Spin(4)\) has two inequivalent ones.

\(Spin(5)\) is isomorphic to \(USp(4)\). We have already seen this fact on the \(B_2=C_2\) Dynkin diagram. The symplectic group is really the unitary group over quaternions, as argued above, so the spinor of \(Spin(5)\) is pseudoreal.

\(Spin(6)\) is isomorphic to \(SU(4)\) which happens to be a proper unitary group from the sequence that has complex representations. So much like for \(Spin(2)\), we have two complex chiral spinor representations that are complex conjugate to each other.

If you combine them, you get a Dirac spinor for \(Spin(6)\) which may also be used as a representation for \(Spin(7)\), using the new matrix \(\Gamma_7\) analogous to \(\Gamma_5\) discussed above. Because this Dirac spinor is \({\bf 4}\oplus \overline{\bf 4}\), it is a real representation: the structure map simply exchanges the two terms which are complex conjugate to each other. That's why the single spinor of \(Spin(7)\) is a real representation.

Finally, \(Spin(8)\) has two inequivalent real chiral spinor representations. Here it would be too singular to consider \(Spin(0)\) but the result you might guess would be right. Why are the spinor representations of \(Spin(8)\) real? It's because the \(D_4=SO(8)\) Dynkin diagram – the logo of Mercedes-Benz of a sort – has some special property, a true gem. It has not only a \(\ZZ_2\) symmetry but a bigger one, \(S_3\), the permutation of all the three Mercedes-Benz external legs. This symmetry is known as the triality.

This triality is actually able to exchange not two irreducible reprentations but it permutes triplets of representations. In particular, you may find out that there are three inequivalent 8-dimensional representations of \(Spin(8)\): the 8-dimensional real vector and two chiral 8-dimensional spinor representations. Note that the Dirac spinor has to have \(2^{8/2}=16\) components. But because there is a triality symmetry, the 8-dimensional representations must come in a triplet of equal types. And because the vector representation is manifestly real, the two chiral spinor representations must be real 8-dimensional representations, too!

The orthogonal groups \(Spin(9)\) and higher are no longer isomorphic to any "simpler" \(SU(m+1)\) or \(USp(2m)\) groups but the reality types are the same as the reality types of their siblings who are younger by 8 dimensions.

The last thing I want to mention is that if you replace \(SO(m+n)\) by \(SO(m,n)\) i.e. if you flip the signature and make some of the dimensions time-like, the dimensions of the spinor representations (considered as representations over \(\CC\), as always) don't change: the dimension of the spinor is the same as it was for \(Spin(m+n)\). However, the reality type of the representation generally does.

The rule is that the reality type is only affected by \((m-n)\). For example, M-theory has a \(Spin(10,1)\) symmetry in \(d=11\). You may see that the spinor is \(32\)-dimensional, much like the Dirac spinor in 10 dimensions regardless of the signature. There is only one inequivalent spinor because the M-theoretical spacetime dimension is odd. And is it real, complex, or pseudoreal? The difference is \(10-1=9\) so the reality is just like for \(Spin(9)\) which is, like for \(Spin(1)\), producing one real spinor. So M-theory offers a 32-dimensional real spinor: that's how its minimal (and simultaneously maximal) set of 32 real supercharges, \(\NNN=1\) in \(d=11\), transforms.

The reality types of the fundamental (and other) representations change for non-compact versions of the other groups, too: unitary, symplectic, exceptional.

Oooh wow, is this for me ... :-D ?

ReplyDeleteIt is exactly the issue from the previous QM article I need some additional explanation about ...

Thanks anyway :-)

Yup, mostly for you, although I hope that if you find the text repulsive enough, it may end up being for someone else, too. ;-) Among the few thousands of viewers who separately open this blog entry at some point, I guess that only a few dozens will read every sentence - others will find it either too trivial or too impenetrable.

ReplyDeleteI'll see shortly ;-), just finishing reading a chapter in one of my Demystified books before ...

ReplyDeleteYou know, I am very stubborn, persistent, and obstinate up to a real pain in the neck if needed :-P

Cheers!

Now I like and appreciated this text a lot, I think it is an as gentle as possible introcuction to representation theory as there can be in the context of such a blog article :-).

ReplyDeleteMost parts of it I could at least follow and I know to feel really completely comfortable and confident about all these things, some more than just reading TRF articles is needed...

So I guess the ball is now on my side of the playing field :-D

Cheers

Wow...Amazing post Lubos! Just a question, under what conditions a group representation can be equivalent to its "complex conjugate" representation (in the case of "complex groups").

ReplyDeleteHi! Thanks! I was using the term "complex conjugation" with two different meanings at different points. In one of them, it was a linear representation over C, i.e. one that allows to multiply by C.

ReplyDeleteThen, when I assumed that the reader already got used to the fact that everything is over C by default, I renamed "complex representation" to what it's actually used for: it's one that isn't equivalent to its complex conjugate representation. It's equivalent to the absence of the "j" structure map because "j" is exactly the map that defines the equivalence between a representation and its complex conjugate!

So complex reps, by definition, can't be equivalent to their complex conjugates. Note that pseudoreal (quaternionic ones) are equivalent to their complex conjugate.

I appreciate these posts as an interested amateur, Luboš.. I never got past topology in my math studies, but it's neat getting some gist of what you guys are playing with. Thanks.

ReplyDeleteVery interesting memo.

ReplyDeleteI wonder which practical - useful for physics - relations exist between all these kind of representations (real, complex, quaternionic) and type of manifolds (Riemann, Kahler, Hyperkahler).

And more precisely, if complex are so fundamental (for instance for Quantum Mechanics), why don't we use Kahler manifolds as base spaces (for instance for gravitation).

Hi! Generic Riemann surfaces are linked to real numbers, Kahler to complex numbers, and hyper-Kahler to quaternions, of course.

ReplyDeleteHowever, by taking about spacetime manifolds, you're missing the point. The spacetime manifold is one of the things that is *not* fundamental. The previous sentence may be debatable but what's not debatable is that the spacetime manifold has nothing to do with quantum mechanics' being the underlying theory. Spacetimes exist in classical theories, too.

Observables such as positions and velocities are routinely given by Hermitian operators so there is a reality condition on them. However and again, this reality/Hermiticity condition is always imposed at the very end. A priori, one must start with generic operators. Products of two operators - even if they're Hermitian - is a general complex operator, not necessarily Hermitian or anti-Hermitian. The condition that an operator is Hermitian (or normal which allows it to be unitary etc.) is what needs to be imposed for the operator to represent an observable.

So the observables such as positions of particles on compactification manifolds have a natural real/Hermiticity condition imposed on them, but this restriction is done at the very end, not at the beginning. There's nothing wrong about real manifolds because they're just some technical aspects of the classical theory, one that also gets imprinted to the quantum theory, not a fundamental thing.

With this being said, it's still true that Kahler manifolds surely do play a preferred role. A vast majority of compactification manifolds that are considered in string theory are Kahler manifolds. The fact that the holonomy is in U(N) - well, SU(N) for Calabi-Yaus - is linked to the conservation of supersymmetry which is indeed a natural thing, too. The most natural/simple vacua are those with the maximum number of SUSIES and they allow the maximum number of "complex-like structures". On the contrary, if you need to get to the "generic real manifolds" etc., you need to add lots of symmetry breaking and extra structure. So in this sense, this intuition that "the fundamental nature of complex numbers should mean that one deals primarily with Kahler manifolds, even for the spacetime compactifications" is valid as well.

Also, the complexification is important in actual calculations. The Wick rotation, Euclidean spacetimes, thermal quantum field theory, and analytic continuations of black holes and other things are examples of the great power of complexified spacetime coordinates in situations where the "natural" character of the spacetime coordinates is assumed to be real.

One must carefully distinguish these complexified spacetimes from the previous Kahler spacetimes - like Calabi-Yau spaces used for compactifications. The difference is that in complexified spacetimes, the actual physics deals with real manifolds. The values of fields in complex values of the coordinates are given by analytic continuations - they're assumed to be holomorphic functions of these coordinates.

That's not true e.g. for the Calabi-Yau (Kahler) manifolds in string theory. While they carry a complex structure on them, the fields are still allowed to depend on these complex coordinates on a Calabi-Yau or another Kahler manifold non-holomorphically.

I recall with horror the math course I took at university called "Representation theory for groups". The professor was nice to me and let me pass, but at that point I knew I had reached my limit. So I shifted my focus to more profitable subjects...

ReplyDeletePerhaps I was unfortunaed in the formulation of my question. But you answered the question anyway, I did a quite elementary introduction to group theory in my blog, and I am aiming to to a second advanced one in the future, speaking about classical groups, spin groups and likely supergroups. You have really excellent blog-posts about group theory disseminated, have you thought in reformatting them in order to convert them into an informal thread on group theory topics! I think it should not be too much effort to make a category for them ;). They would be availabre easily instead of that, and I am sure that even you would benefit from your own wonderful materials. :).

ReplyDelete