Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a compact Lie group and let $R(G)$ denote its complex representation ring. If $G$ is simply connected, such as $G_2$, $F_4$ or $E_8$, then it is known that $R(G)$ is a polynomial ring [F. Adams, Lectures on Lie groups ('69), Theorem 6.41].

Question: Are the representation rings of the adjoint groups $E_6$ and $E_7$ polynomial rings as well?

(Edit: the following articles are irrelevant as they consider the simply connected cover rather than the adjoint group.) In [T. Watanabe, The Chern Character of the Symmetric Space EI, Publ. RIMS ('95)] it is claimed that it is shown in [H. Minami, K-Groups of Symmetric Spaces I+II, Osaka J. Math. (75'+76')] that $R(E_6)$ is a polynomial ring. I wasn't able to locate this precise claim in the second source, though. I've found no reference concerning $R(E_7)$.

share|improve this question
1  
There are two different compact Lie groups of type $E_6$ (simply connected and adjoint ones), so the symbol $R(E_6)$ is not quite well defined. For any simply connected compact group $G$ the ring $R(G)$ is a polynomial ring (with generators given by the fundamental representations). –  Victor Ostrik Jul 10 '13 at 21:08
    
It looked to me from wikipedia that there is only one compact (real) form of $E_6$. This has $\pi_1(E_6)\cong\mathbb Z/3$. –  Rasmus Bentmann Jul 10 '13 at 21:24
3  
For every compact Lie algebra (i.e. real Lie algebra with negative-definite Killing form), there are two distinguished compact groups: the adjoint group and its simply-connected cover. The relationship is that pi_1(adjoint) = center(simply-connected). These two forms might be the same (as in the case of E_8), and there might also be other forms between them (SL(4) > SO(6) > PSL(4)). This is what @VictorOstrik was referencing. Perhaps you should clarify that you care about the adjoint forms? –  Theo Johnson-Freyd Jul 10 '13 at 21:43
    
@TheoJohnson-Freyd: Thanks for explaining. Will edit to clarify. (Edit: done.) –  Rasmus Bentmann Jul 10 '13 at 21:46

3 Answers 3

up vote 9 down vote accepted

As Victor points out in his comment, your formulation is somewhat out of focus. The general fact is that the representation ring of a connected semisimple compact Lie group is a polynomial algebra iff each simple factor of the group is simply connected or is an odd dimensional special orthogonal group. It's useful to consult the treatment in the Springer GTM 98 Representations of Compact Lie Groups by Brocker and tom Dieck (pages 255-256) along with the references given there. I've checked for example the useful paper by Steinberg referred to there.

So for exceptional groups of type $E_6, E_7$ which are of adjoint rather than simply connected type, the representation ring fails to be a polynomial ring (but is always finite generated).

P.S. Concerning notation, you should write $R(G)$ for the representation ring of a given compact Lie group, since this can vary with the isogeny type in a fixed Lie type such as $E_6, E_7$. The general result found in references like those I mentioned is that $R(G)$ is isomorphic to the ring of invariants of the Weyl group $W$ of a maximal torus $T$ in the integral group ring of the character group $X(T)$. For an adjoint group $G$ this is the root lattice rather than the full weight lattice of the root system, which can complicate the description a lot. I'm not aware of explicit formulas in the literature for your special cases, but classical types appear in some Bourbaki exercises (see Chap. VI, $\S3$ and Chap. IX, $\S7$ of Lie Groups and Lie Algebras).

One small example: In type $A_1$, take $G= \mathrm{SU}(2)$ (simply connected) with $X(T)$ spanned by fundamental weight $\varpi$ and $W$ of order 2. Formally setting $x = e^\varpi$, we get $\mathbb{Z}[X(T)] = \mathbb{Z}[x,x^{-1}]$ (Laurent polynomials) with a reflection in $W$ sending $x \mapsto x^{-1}$; so $R(G) \cong \mathbb{Z}[x+x^{-1}]$.Taking $G= \mathrm{SO}(3)$ (adjoint), the root `$\alpha = 2\varpi$ spans $X(T)$, so instead $R(G) \cong \mathbb{Z}[x^2 + x^{-2}]$. (Since $A_1 = B_1$, the ring is polynomial in both cases.)

share|improve this answer
    
Thank you very much for your answer! I would still be interested in explicit descriptions of the rings $R(E_6)$ and $R(E_7)$. Can they be described as quotients of polynomial rings, like $R(SO(2n))$? –  Rasmus Bentmann Jul 10 '13 at 22:10
    
My previous comment is addressed in @ARupinski's answer. –  Rasmus Bentmann Jul 11 '13 at 14:10

This started as a comment in response to the followup by the OP to Jim's answer, but quickly grew to span several comment blocks, so I'm just expanding it into an answer block.

"I would still be interested in explicit descriptions of the rings $R(E_6)$ and $R(E_7)$. Can they be described as quotients of polynomial rings, like $R(SO(2n))$?"

The short answer is: yes, we can get explicit descriptions, in theory at least.

If $\tilde{G}$ denotes a simply connected group and $G$ its adjoint group, then the irreps of $\tilde{G}$ have a natural grading by the elements of $Z(\tilde{G})$ given by looking at the image of the center in the given irrep. This grading is preserved by taking tensor products, so in particular we can find a set of irreps which span the representation ring of the adjoint group. Irreps of the adjoint group are those one which this grading is trivial; there are then relations among these irreps which can be obtained by looking in the representation ring of $\tilde{G}$.

For example, let $\tilde{G}=SU(3)$ with fundamental weights $w_1$ and $w_2$. In this case an admissible $\mathbb{Z}/3\mathbb{Z}$ grading on an irrep $\pi_{a\cdot w_1+b\cdot w_2}$ (the irrep with highest weight $a\cdot w_1+b\cdot w_2$) of $SU(3)$ is given by $(a−b)\mod 3$. One then sees that all irreps of $G = SU(3)/3$ are generated by $\pi_{3w_1}$, $\pi_{3w_2}$, and $\pi_{w_1+w_2}$; but as $SU(3)/3$ has rank 2, there must be a relation among these three generators to reduce the "dimension" (for whatever the correct notion of dimension of a ring is appropriate here) of its representation ring down by 1; i.e. the representation ring is a quotient of a polynomial ring.

The exact relation can be found by considering tensor products of these three generators and subtracting off appropriate highest weight pieces until one reaches 0. So one starts with $\pi_{w_1+w_2}^{\otimes 3}$ and subtracts off $\pi_{3w_1}\otimes \pi_{3w_2}$ then looks at the highest weights of this representation and subtracts off an appropriate tensor product of the three generator irreps to cancel the highest weights left and repeats until one reaches 0; the resulting polynomial relation has the form (unfortunately I don't recall the exact coefficients $n_i$ and do not currently have access to my Maple worksheets where they are stored):

$\pi_{w_1+w_2}^{\otimes 3} - \pi_{3w_1}\otimes \pi_{3w_2} + n_1\cdot(\pi_{3w_1}\otimes\pi_{w_1+w_2} + \pi_{3w_2}\otimes\pi_{w_1+w_2}) + n_2\cdot \pi_{w_1+w_2}^{\otimes 2} + n_3\cdot(\pi_{3w_1} + \pi_{3w_2}) + n_4\cdot\pi_{w_1+w_2} + n_5\cdot 1 = 0$

Note that this relation is not a problem in $R(SU(3))$ (which is a polynomial ring) since there $\pi_{3w_1}$, $\pi_{3w_2}$, and $\pi_{w_1+w_2}$ are themselves polynomials in the fundamental representations $\pi_{w_1}$ and $\pi_{w_2}$; it is the failure of these two fundamental representations to live in $R(SU(3)/3)$ that forces us to take a 3-element set of generators of $R(SU(3)/3)$ and also forces this 3-element set to have a single relation.

In application to your question, you would need to do a similar construction in $R(E_6)$ and $R(E_7)$: start with the grading function applied to irreps to find a set of irreps of $\tilde{E}_6$ or $\tilde{E}_7$ which generate each subring; then work out relations among these generators by looking at the highest weights and adding/subtracting off appropriate tensor powers of these generators until one arrives at 0. As I don't recall the grading function for these two groups offhand, I couldn't tell you how many generators to expect for each, but once you have all $g_6$ generators for $R(E_6)$, you ought to be able to find exactly $g_6 - 6$ independent relations among them and similarly for the generators of $R(E_7)$.

Update: Since it is not difficult to compute the grading function, I went ahead and worked them out so that I could elaborate. The $Z(\tilde{G})$ grade of an irrep $\pi_\omega$ of $\tilde{G}$ is trivial $iff$ $\pi_\omega$ contains the zero weight and the $Z(\tilde{G})$ grades of two irreps are equal iff the highest weight of one is contained in the weights of the other (depending of course on which weight dominates which).

Equivalently if one can reach the zero weight from $\omega$ by subtracting off fundamental roots, or equivalently $\omega$ is a sum of fundamental roots then the grade of $\pi_\omega$ is zero. Another way to think about it is that the grading is really describing the image of the weight in the quotient of the weight lattice modulo the root lattice.

Since the Cartan matrix expresses weights as sums of fundamental roots, we can use it to solve for this image for an arbitrary weight; simply invert the Cartan matrix and multiply it by the given weight (expressed as a vector over the basis of fundamental weights). If all the entries of the resulting vector are integers, then the weight has trivial grading. If not all entries are integral, then the weight has a nontrivial grade (and there is some choice to this; the choice is restricted by the condition that two weights have the same grade if one dominates the other and the different grading functions can be obtained from one another via a certain action of the outer automorphisms of $\tilde{G}$, so they are all essentially equivalent).

Using the Bourbaki numbering (as found on https://en.wikipedia.org/wiki/Root_system for example), one obtains for $E_6$ that one admissible weight function for $\pi_{\sum_{i=1}^6n_iw_i}$ is given by:

$(n_1+2n_2+n_5+2n_6)\mod 3$

For $E_7$, the weight function for $\pi_{\sum_{i=1}^7n_iw_i}$ is given by:

$(n_1+n_3+n_5)\mod 2$

Now from these we can recover the necessary generators for our adjoint representation rings: we need to find a set of highest weights with trivial grade which integrally span the trivially graded part of the weight lattice (= the root lattice).

For $E_6$ you can check that this set is:

$\{w_3,w_4,3w_1,3w_2,3w_5,3w_6,w_1+w_2,w_1+2w_5,w_1+w_6,w_2+w_5,w_2+2w_6,w_5+w_6\}$

So there are $g_6 = 12$ generators for $R(E_6)$ and so we should be able to find 6 independent polynomial relations among their corresponding irreps which, once known, give the structure of the quotient ring.

For $E_7$ the set of spanning weights is:

$\{w_2,w_4,w_6,w_7,2w_1,2w_3,2w_5,w_1+w_2,w_1+w_3,w_2+w_3\}$

So there are $g_7 = 10$ generators for $R(E_7)$ and so we should be able to find 3 independent polynomial relations among the corresponding irreps.

share|improve this answer
    
Many thanks for your answer! –  Rasmus Bentmann Jul 11 '13 at 6:57

Here is the printable version of a lecture by David Vogan on the character table of the the split real group of type $E_8$. Here are the (non-printable) slides of this talk. A very ingeneous and huge collaboration went into the computer computation of this table, and the resulting table apparently has 60GB.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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