Find a dual lattice basis for lattice $e_1 = \sqrt 2 \hat e_x, \ e_2 = -\frac{1}{\sqrt 2} \hat e_x + \sqrt{\frac{3}{2}}\hat e_y$

Question

Given the basis vector of a lattice $L$:

$$e_1 = \sqrt 2 \hat e_x, \ e_2 = -\frac{1}{\sqrt 2} \hat e_x + \sqrt{\frac{3}{2}}\hat e_y$$

I want to find a set of basis vector for the dual lattice $L^*$.

And the 2nd part of the problem: show that for vector $(p_L,p_R)$, where $p_L, \ p_R \in L^*$ and $p_L-p_R\in L$, if we use the norm $p^2=p^2_L - p^2_R$ (Narain), we have defined a even self-dual lattice.

And there is the last part of the question, if you are interested. Above construction give a special point in moduli space of $T^2$ compactification of bosonic string, with an enhanced $SU(3)\times SU(3)$ symmetry. Show that there are 16 massless gauge bosons at this point.

I am quite new to lattice. Here is what I tried:

It is clear we want to get $\langle e_i^*,e_j \rangle\,\in \mathbb Z \forall i,j$. This give me 4 equation when I plug in $e_j$. Now I can plug in different integer on the left hand side and I will get some basis and span them to get a set of points for each basis.

However, there are two problems. First, I also get parts of original lattice, and since zero vector is always in the result lattice, I will surely get some point that is in original lattice. Also, I may not necessary get all dual lattice, this can be avoid by carefully choose which integer to plug in.

So to avoid this problem I think I have to move my origin by a vector say $\xi=\frac{\sqrt 2}{2} \hat e_x + \frac{\sqrt 6}{6}\hat e_y$. And any point in $L^*$ can be write as $\xi + n e_1 + m e_2$ or $2\xi + n e_1 + m e_2$. But seems this is not what the question wants already.

If I want to carry this to the 2nd part of the problem. First we see $p_L-p_R\in L$ is not always true since I can left a $\xi$ suggesting this is a point of dual lattice (so what I done early should be wrong). Also it is very unclear what the inner product should be for p. Does it mean $(p,q) = \langle p_L,q_L \rangle -\langle p_R,q_R \rangle$? In class we have done some thing relate to this but $p_L$ and $p_R$ were numbers but here they are lattice points which confused me.

The last problem is a physics question which I have some clues but not quite clear. I think I need to find the adjoint representation of $SU(3)\times SU(3)$ (I hope there are 16 generators) and show that the corresponding gauge boson is massless (but if there is no SSB they should be massless from beginning).

Answer 1

I can only help with the first part of the question. When it comes to basis of lattices I find it much easier to do everything in terms of matrices. The basis of L (column vectors) is: $$B=\begin{bmatrix} \sqrt2 & -\frac{1}{\sqrt2} \\ 0 & \frac{\sqrt{3}}{2} \end{bmatrix}$$ The basis of $L^*$ is simply the row vectors of $$B^{-1}=\begin{bmatrix} \frac{1}{\sqrt2} & \frac{1}{\sqrt3} \\ 0 & \frac{2}{\sqrt{3}} \end{bmatrix}$$

I find the second part confusing: If $p_L,p_R\in L^*$ then wouldn't you also have $p_L-p_R\in L^*$ since lattices are closed under addition? Maybe check the wording of the problem again?

Answer 2

After a few emails to my professor, I have solve the question. First, my solution in the question is correct. By solving inner product of basis is integer, we can get a basis like $$e^*_1 = \frac{1}{\sqrt 2} \hat e_x - \frac{1}{\sqrt 6} \hat e_y, \ e^*_2=\sqrt{\frac{2}{3}}\hat e_y$$ The span of this basis do give you $L^*$. My concern is not really a problem since $L \in L^*$ is fine. I should not try to avoid the original lattice by add a $\xi$ vector.

Now we can see that it is reasonable to say $p_L-p_R\in L$ since $L$ is in $L^*$.

Move the the next part of the problem. My understanding is correct. We can follow the inner product in the question and require the integer result. This will give us a new basis for the dual lattice. However, we will found the new basis is the same as the basis given in the problem. So it is self-dual.

For the even part. When calculating $p^2$, we will get $\frac{2}{3}(\text{something})$. But since it is self dual, $p^2$ have to be an integer. So $\frac{(\text{something})}{3}$ have to be an integer. Thus $p^2$ is always even. Thus we have showed it is a even self-dual lattice.

In the class we are talking about a $S^1$ string compactification, but here we are in $T^2$, this is the reason we found $p_L$ and $p_R$ to be a vector.