Saturday, June 01, 2013

Quintuplets in physics

Cool anniversary: In late January, we celebrated the 30th anniversary of the announcement of the discovery of the W-boson. Today, we celebrate the 30th anniversary of the Z-boson. They were comparably important discoveries to the recent discovery of the God particle.

Sport: Viktoria Pilsen defeated Hradec, a much weaker team, 3-to-0 in the last round so we won the top soccer league for the 2nd time (after 2011). Because the Pilsner ice-hockey team has won the top league as well, Pilsen became the 2nd town in Czechia after Prague that collected both titles in the same year (correction: wrong, 3rd town, Ostrava did it in 1981).
Ms Alexandra Kiňová (23) is expecting the first Czechia's naturally born quintuplets (a package of 5 babies) on Sunday morning (tomorrow; update: they're out fine) which would mean that we match the achievement of the most fertile U.S. state – Utah – from the last week.

The Daily Mail tells us that the pregnancy has been easy so far. Doctors were still talking about "twins" in January and "quadruplets" in April. The probability that a birth produces \(n\)-tuplets goes like \(1/90^{n-1}\) or so but the decrease slows down relatively to this formula for really high representations.

In physics, quintuplets are rare, too. By quintuplets, we mean five-dimensional irreducible representations of groups.

Correct me if I am wrong but I think that among the simple Lie groups, only \(SU(2)=SO(3)\), \(USp(4)=SO(5)\), and \(SU(5)\) have irreducible five-dimensional representations. Let's look at them because looking at all quintuplets in group theory and physics is a rather unusual direction of approach to a subset of wisdom contained across the structure of maths and physics.

First, \(SU(2)\). That's a three-dimensional group of \(2\times 2\) complex matrices \(M\) obeying \(MM^\dagger={\bf 1}\) and \(\det M=1\). The basic isomorphisms behind spinors imply that this group is the same as the group \(SO(3)\) of rotations of the three-dimensional space except that the matrices \(+M\) and \(-M\) have to be identified.

The irreducible representations of \(SU(2)\) are labeled by the spin \(j\) which must be either non-negative integer or positive half-integer (only the former may also be interpreted as proper representations of \(SO(3)\); the latter change their sign after a 360-degree rotation). Because the \(z\)-projection goes from \(m=-j\) to \(m=+j\) with the spacing equal to one, the representation is \((2j+1)\)-dimensional.

The \(j=0\) representation is the trivial singlet that doesn't transform at all; the \(j=1/2\) is the two-dimensional pseudoreal spinor; the \(j=1\) representation is equivalent to the usual 3-dimensional vector; the \(j=3/2\) representation is a gravitino-like four-dimensional "spinvector". And finally, the \(j=2\) representation is the traceless symmetric tensor. What do I mean by that?

Imagine that you consider the tensor product \(V\otimes W\) of two copies of the three-dimensional vector space \(V=W=\RR^3\). The tensor product is composed of objects \(T_{ij}\) where \(i,j\) are vector indices: it's composed of tensors. Clearly, such a tensor has \(3\times 3 = 9\) independent components. They can be split into several pieces:\[

{\bf 3}\otimes {\bf 3} = {\bf 5} \oplus {\bf 1}\oplus {\bf 3}

\] The identity \(3\times 3 = 5+1+3\) is the consistency check that verifies that the representations above have the right dimensions but the boldface identity above says more than just the arithmetic claim about the integers: the two sides are representations of whole groups and the identity says that they're transforming in equivalent ways under all elements of the group. Why is this decomposition right? Well, the tensor \(T_{ij}\) may be divided to the symmetric tensor part which is 6-dimensional and the antisymmetric tensor which is 3-dimensional (it is equal to \(\epsilon_{ijk}v_k\) i.e. equivalent to some vector \(v_k\)).

However, the 6-dimensional symmetric tensor isn't an irreducible representation of \(SO(3)\). The trace \[

\sum_{i=1}^3 T_{ii}

\] is independent of the coordinate system i.e. invariant under rotations and may be separated from the 6-dimensional representation. The trace may be set to zero by removing it i.e. considering\[

T^\text{traceless part}_{ij} = T_{ij} - \frac 13 \delta_{ij} T_{kk}

\] and such a traceless tensor has 5 independent components; it is a quintuplet. The quadrupole moment tensor is one of the most famous applications of this 5-dimensional object. You could think it's just an accident that this number 5 is equal to the number of integers between \(m=-2\) and \(m=+2\); you could claim that the agreement is pure numerology, an agreement between the dimensions of two representations. But it is more than numerology: the representations are completely equivalent. The translation from the components \(T_{ij}\) of the (complexified) traceless tensor and the five complex amplitudes \(c_m\) for \(-2\leq m\leq 2\) is nothing else than a linear change of the basis. It has to be so because for every \(j\), the representation of \(SU(2)\) is unique.

Now, let's talk about \(SO(5)\). Clearly, this group of rotations of the 5-dimensional space has a 5-dimensional vector representation consisting of \(v_i\). But what some readers aren't aware of is that the group \(SO(5)\) may also be identified with the isomorphic \(\ZZ_2\) quotient of a spinor-based group, namely \(USp(4)\). What is this group? It's a unitary (U) symplectic (Sp) group of complex \(4\times 4\) matrices \(M\) that obey\[

MM^\dagger = M^\dagger M = 1, \quad M A M^T = A.

\] Both conditions have to be satisfied. The first condition is the well-known unitarity condition, effectively meaning that \(s_i^* s_i\) is kept invariant (it's the squared Pythagorean length of the vector computed with the absolute values). The other condition is equivalent to keeping the antisymmetric cross-like product of two vector-like objects \(s_i A_{ij} t_j\) invariant where \(A_{ij}\) are elements of the (non-singular) antisymmetric matrix \(A\) above. Note that in this invariant, there is no complex conjugation.

Simple linear redefinitions of the 4 complex components \(s_i\) may always translate your convention for \(A\) mine which is \[

A = \text{block-diag} \zav{ \pmatrix{0&+1\\-1&0}, \pmatrix{0&+1\\-1&0} }

\] You just arrange the right number of the "simplest nonzero antisymmetric matrices" along the (block) diagonal. The two conditions (unitary and symplectic) may be then seen to imply that \(M\) is composed of \(2\times 2\) blocks of this form\[

\pmatrix{ \alpha&+\beta\\ -\beta^*&\alpha^*},\quad \alpha,\beta\in\CC

\] and the addition+matrix-multiplication rules for such matrices are the same rules as the addition+multiplication rules for the quaternions \(\HHH\). So the group \(USp(2N)\) may also be called \(U(N,\HHH)\), the unitary group over quaternions. In particular, \(USp(4)=U(2,\HHH)\). Such a quaternionization is possible with all pseudoreal representations.

So the 4-dimensional complex (actually pseudoreal!) fundamental representation of \(USp(4)\) is complex-4-dimensional (but it is equivalent to its complex conjugate because it's pseudoreal!) and it may be viewed as a spinor of \(SO(5)\). It is no coincidence that \(4\) in \(USp(4)\) is a power of two. How do you get the five-dimensional \(j=1\) vector out of these four-dimensional spinors?

Note that for \(SO(3)\sim SU(2)\), we had\[

{\bf 2}\otimes{\bf 2} = {\bf 3}\oplus {\bf 1}.

\] The tensor product of two spinors produced a vector (triplet; also the symmetric part of the tensor with two spinor indices) and a singlet (the antisymmetric part of the tensor with two 2-valued indices). Similarly, here we have\[

{\bf 4}\otimes{\bf 4} = {\bf 5}\oplus {\bf 1}\oplus {\bf 10}.

\] The decomposition of \(4\times 4 = 16\) to \(6+10\) is the usual decomposition of a "tensor with two spinor indices" to the antisymmetric part and the symmetric part, respectively. The symmetric part may be identified as the antisymmetric tensor with two vector indices, note that \(5\times 4 / 2\times 1 = 10\). And the antisymmetric part is actually irreducible here. It's because the invariant for the symplectic groups is antisymmetric, \(a_{ij}\), rather than the symmetric \(\delta_{ij}\) we had for the orthogonal groups, so it's the antisymmetric part that decomposes into two irreducible pieces.

By tensor multiplying \({\bf 4}\) with copies of itself, we may obtain all representations of \(USp(4)\) and \(SO(5)\) by picking pieces of the decomposed tensor products. That's what we mean by saying that the representation \({\bf 4}\) is "fundamental". Whenever an even number of these \({\bf 4}\) factors appears in the tensor product, we obtain honest representations of \(SO(5)\) that are invariant under 360-degree rotations and all these representations may also be given a natural description in terms of tensors with vector indices.

Finally, the special unitary group \(SU(5)\) has an obvious 5-dimensional complex representation. It is a genuinely complex one, i.e. a representation inequivalent to its complex conjugate:\[

{\bf 5}\neq \overline{\bf 5}

\] This representation (and its complex conjugate, of course) is important in the simplest grand unified models in particle physics. One may say that \(SU(5)\) is an obvious extension of the QCD colorful group \(SU(3)\). We keep the first three colors (red, green, blue, so to say) and add two more colors that are interpreted as two lepton species from the same generation. The full collection of fifteen 2-component left-handed spinors per generation (they describe quarks and leptons; a Dirac spinor is composed of two 2-component spinors; the right-handed neutrino is not included among the fifteen) is interpreted as \[

{\bf 5}\oplus\overline{\bf 10},

\] the direct sum of the fundamental quintuplet of \(SU(5)\) we have already mentioned and the antisymmetric "tensor" with \(5\times 4 / 2\times 1\) components. Note that the counting of the components is the same as it was for the representation of \(SO(5)\) above. However, the 10-dimensional representation of \(SU(5)\) is a complex one, inequivalent to its complex conjugate (I won't explain why the bar appears in the decomposition above, it's a technicality). The list of 15 spinors may be extended to 16, \(10+5+1\), if we add one right-handed neutrino and this \({\bf 16}\) is then the spinor representation of \(SO(10)\), a somewhat larger group that is capable of being the grand unified group (it is no accident that 16 is a power of two: that's what spinors always do).

The number 5 may be thought of as the first "irregular" integer of a sort but it is still small and special enough and is therefore linked to many special things in maths and physics. In maths, five is special because the square root of five appears in the golden ratio; and a pentagram may be constructed by a pair of compasses and a ruler (these two facts are actually related). Quadrupole moments, moments of inertia, five-dimensional rotations, and grand unifications are among the physical topics in which 5-dimensional representations are used as "elementary building blocks".

I hope that Ms Kiňová's birth will be as smooth as her pregnancy.


  1. Wow she could make a lot of money if she were to sell the babies to gay couples... Is it allowed in Tchéquie ? (I'm being sarcastic although this could be the case).

  2. I am sure she will be doing fine without that, too. Ministers etc. are talking about it, effectively preparing an Amendment of the Constitution about the Quintuplets, if I exaggerate just a little bit - extra aid etc.

  3. Look at this cute and funny video of quadruplets laughing :-) (and then let's imagine when they cry all together).

  4. That's cute! LOL.

  5. No , she will give only one baby because there's $Z_{5}$ symmetry between them which give us the same lagrangian .

  6. Would you call that a litter?

  7. Alejandro RiveroJun 1, 2013, 4:00:00 PM

    One of my silly questions in MO was about SO(5)

  8. I totally agree with you. It is silly. How can SU(3) be SO(5)? SU(3) is 8-dimensional, SO(5) is 10-dimensional. The dimensions are different but so close that clearly none of the two may be a subgroup of the other.

  9. Thanks for this nice representation of some group theory Lumo ;-)

    Do you know in a bit more detail what is inside Nadir Jeevanjee's book? Does it give nice cool physics examples as you always do, while explaining group theory?

  10. That's definitely most endearing quads and a nice mum.
    But when I look at the photo of the very happy and glowing-looking face of the quintuplets-expecting woman - i.e. a female who is in actual fact in a very precarious state with five parasites growing inside her so that they almost make her belly burst, then it makes me think: "That's how strongly and ruthlessly Mother Nature beats her biological (bongo-) drum! %-|

  11. Something tells me there is an exponential sign tattooed on her belly.

  12. Dilaton , If you want to learn some group theory , read georgi or brian hall . They are very good and straightforward to read.

  13. Wow, that's more babies than the Germans have in a whole year. The strategic implications are obvious. It's the first step towards Czech hegemony in Europe!