# relating spinor and fundamental representation for $E_8$

While proving a very important relation which is satisfied both by $SO(32)$ AND $E_8$, which makes it possible to factorize the anomaly into two parts. The relation is $Tr(F^6)=\frac{1}{48}TrF^2TrF^4-\frac{1}{14400}(TrF^2)^3$, where trace is in adjoint representation.

I am able to prove this relation but while doing so, I have some identities which relates the spinor representation $128$ of $SO(16)$ to fundamental representation of $SO(16)$ which I must show but this is not working out.

The simplest one being $TrF^2=16trF^2$, where $Tr$ is in spinor representation $128$ and $tr$ is in fundamental representation. There are other relations showing the equality between $TrF^4$ and $tr(F^2)^2$ and $trF^4$. I am aware that spinor representation would be $\sigma_{ij}$ which is $128$ dimensional. While trying to prove these identities, I have noticed that if $F^2$ in the fundamental representation is diagonal with only two elements -1 and -1 and if $\sigma_{ij}^2=\frac{-I}{4}$ where $I$ is $128$ dimensional identity matrix then I can get the result. But I can not convince myself why it should be true.

Any details would be appreciated of how to prove it. The identities can be found in GSW chapter 13 last section (VOL.2).

-
Note that, for $E8$, you have $Tr(F^4)=36(\frac{1}{60} Tr(F^2))^2$ and $Tr(F^6)=30(\frac{1}{60} Tr(F^2))^3$, cf formulae $(176), (177) p. 35$ of this paper. Your first equality then is easily checked. –  Trimok Aug 2 at 10:36
I wonder if you may use the formulae $(65), (66), p. 19,20$ of this same paper to establish the relations between traces in the spinor representation and traces in the adjoint representation (of $SO(16)$). –  Trimok Aug 2 at 10:46
Thanks for the reference but formula $Tr(F^4)=36(\frac{1}{60} Tr(F^2))^2$ and the other one is precisely what I want to prove by decomposing the adjoint representation of $E_8$ as 248=120+128, I can carry out the trace in 120 representation from a formula which I asked earlier on this forum( and hence proved). It is the 128 representation which is giving me problem. The formula 65 and 66 looking more difficult to verify by any means , I guess there should be an easy way of doing it. Thanks for the reference though. –  user44895 Aug 2 at 11:37
Ok, I am able to verify the identities relating traces in $128$ representation to vector representation using formula $65$ and $66$ of the reference and hence I can complete the proof now. However the formula $65$ and $66$ is still foreign to me, can any one provide some reference regarding how to obtain formula 65 and 66 of the reference cited by Trimok. The reference 26 in the paper which is supposed to deal with it is not in my reach. Thanks for any help. –  user44895 Aug 3 at 7:28
I would be interested if you can provide a complete detailed answer (to your own question), if you have time. Sure I will learn something interesting. –  Trimok Aug 3 at 11:07

Ok, I will sketch a some important steps here for relating spinor representation of $D_n$ to vector representation.

For $E_8$, the adjoint representation 248 can be written in terms of positive chirality spinor representation 128 of SO(16) and adjoint representation 120 of SO(16) as, $248=120+128$

Now carrying out the trace in 120 representation can be performed using Chern character factorization property which I asked earlier. The result which relate trace in adjoint representation to fundamental representation reads,

$Tr(e^{iF})= \frac{1}{2}(tre^{iF})^2-\frac{1}{2}(tre^{2iF})$

Now we have to evaluate the trace in representation 128 and relate it to fundamental representation , so that we can compare the relationship between $TrF^4$ ( adjoint of $E_8$) and $TrF^2$ and $Tr F^6$ to $TrF^2$. These can be related because the independent invariant tensor of E_8 are 2,8, 12 and so on( see ref.1). Now one has to obtain the relation between spinor representation 128 and fundamental representation. From the reference (1) the relevant formula is 65 and 66.

Now $X_r$ is a polynomial of order 8 here and hence we can neglect the second term for our purposes i.e. upto order $TrF^6$.($B_{2n}$ are bernoulli numbers) Now putting $r=8, B_2=1/6$ and $B_4=-1/30$ and writing the left side as $Tre^{F}$ (in 128 representation) , and doing the expansion we can collect term by comparing the powers. One will get by doing this,

$TrF^2= 16trF^2$ and $TrF^4= 6(trF^2)^2-8trF^4$ which are the ones precisely needed. Summing the results from 128 and 120 in terms of fundamental represntation one immediately get $TrF^4= 6(trF^2)^2-8trF^4+8trF^4+3(trF^2)^2=9(trF^2)^2$ and $TrF^2=16 trF^2+14 trF^2=30 trF^2$. It is clear that we have relation,

$TrF^4= (TrF^2)^2/100$

Similarly by including a next power and value$B_6$, we can get the relation.

$TrF^4= (TrF^2)^3/7200$

Which are needed.

Now how to show the equivalence of the big formula 65 and 66 relating the spinor to vector representation can be found in ref. 2.

Now to show the equivalence, one has to note that $F/2\pi$ in fundamental representation can be diagonalised with eigenvalues $\pm iy_{\beta}$ ( since it is an antisymmetric matrix), where $\beta$ =1…….r, however in the spinor representation it’s eigenvalues are $1/2 (\pm iy_1,………..,\pm iy_r)$, where there is an even or odd number of minus signs, according to positive or negative chirality.

This form of eigenvalues is result of expressing the maximal commuting set of generator in tems of it’s weights. I cannot put much emphasis on it here, for those who are intereseted can consult Polchinski Vol.2, chapter 11 or/and Wybourne ‘ classical groups for physicist’. Now we look at formula 3.11 of reference 2,

$Tre^{iF_s/2\pi}=2^{m-1}[\pi_{1}^{m}coshy_\beta/2 \pm \pi_{1}^{m}sinhy_\beta/2]$

. To verify that this formula holds and no cross term like sin.cos is present, we take m=4 case and chose the condition of an odd number of minus sign. The possible combination we have now is,$( y_1,y_2,y_3, -y_4),(y_1,-y_2,y_3,y_4),(y_1,y_2,-y_3,y_4),(y_1,y_2,y_3,-y_4),(-y_1,-y_2,-y_3,y_4),(-y_1,-y_2,y_3,-y_4),(-y_1,y_2,-y_3,-y_4),(y_1,-y_2,-y_3,-y_4)$ with a factor of i/2 removed, Now we evaluate $e^{iF_s/2\pi}$, first we can observe that $(y_1,y_2,y_3, -y_4)$ is supplemented with $(-y_1,-y_2,-y_3, y_4)$ and hence for other term. So we have terms like $e^{\frac{1}{2}( y_1+y_2+y_3 -y_4)}$ and $e^{-\frac{1}{2}( y_1+y_2+y_3 -y_4)}$, which can be combined to give cosh type term. Same is true for the other three combination.

Combining all terms together we get,

$4[cosh(y_1-y_2)cosh(y_3+y_4)+ cosh(y_1+y_2)cosh(y_3-y_4)]$, evaluating it does give

$8[cosh(y_1/2)……cosh(y_4/2)- sinh(y_1/2)……sinh(y_4/2)]$ which is the result following from 3.11 of reference 2.

Even case can be verified in the same manner and any other order.

Now we can use the taylor expansion of $log cosh x$ which can be found in the books which deal with mathematical tables like reference 3 and arrive at first term of 65 and 66 of reference 1.

The second term can also be written in a similar but not same way and it involves a term Which is of order $Tr(F^m)$ and higher (Reference 2, eqn.3.13 and 3.14). However this term does not have it’s use in our case and hence we omit any discussion of it.

References-

1- Ritbergen,Schellekens, Vermaseren,Group theory factors for Feynman diagrams: hep-ph/9802376v1.

2- A.N. Schellekens and N.P. Warner, Nucl. Phys B287 (1987) 317 ( can be found on google scholar).

3- Gradshteyn, Ryzhik, Jeffrey’s book on mathematical tables. (I don’t have the exact name of the book!).

-
+1 for the work. –  Trimok Aug 6 at 10:45