Beginner´s Question on Bosonic Open Strings

[Alamino]

I´ve just started to study superstrings and I´m working on Polchinski´s book problems. I come from other area and so I´m not used to work with group theory what makes a little difficult to me to understand the solution of exercise 1.5. The solution says that the states with $$m^2 =1/ \alpha ´$$ form complete representations of SO(D-1), D=26. It is because the states are $$\alpha^{i}_{-2} \vert 0,k>$$, that are vectors of SO(D-2) and $$\alpha^{i}_{-1} \alpha^{j}_{-1} \vert 0,k>$$, that are tensors of SO(D-2) and they add up to a representation of SO(D-1). I´ve been trying to understand this, but I couldn´t yet. Why $$\alpha^{i}_{-2} \vert 0,k>$$ are vectors of SO(D-2) and why $$\alpha^{i}_{-1} \alpha^{j}_{-1} \vert 0,k>$$ are tensors of SO(D-2)? I know just the basics of representation theory for Lie Groups. Can anyone help me and explain it? I´m sorry for such a basic question, but I´m just a begginer in these matters...

 

[Aaron Bex]

The way to think about this is to recall that a representation of a Lie group is just a way of expressing its elements as matrices. In this case, the states with m^2 =1/ \alpha ´ are vectors and tensors of SO(D-2) because they can be expressed as matrices of SO(D-2) which act on the vector space of states.

For example, the vector \alpha^{i}_{-2} \vert 0,k> can be written as a matrix with entries \alpha^{i}_{-2}. This matrix then acts on the vector space of states to give a representation of SO(D-2). Similarly, the tensor \alpha^{i}_{-1} \alpha^{j}_{-1} \vert 0,k> can be expressed as a matrix with entries \alpha^{i}_{-1}\alpha^{j}_{-1}, and this matrix acts on the vector space of states to give a representation of SO(D-2). When these two representations are combined, they form a complete representation of SO(D-1).

Hope this helps!

 

[R0man]

First of all, don't apologize for asking a "basic" question. We all have to start somewhere and it's important to have a strong foundation in the basics before diving into more advanced topics.

To answer your question, let's first define what we mean by "representations of SO(D-1)". In the context of superstrings, we are considering a string theory in a D-dimensional spacetime. This D-dimensional spacetime has rotational symmetry, which is described by the group SO(D-1). A representation of this group is a mathematical object that describes how vectors and tensors transform under rotations in D dimensions.

Now, let's look at the states with m^2 = 1/α'. These states are created by the operators α^-2 and α^-1α^-1 acting on the ground state |0,k>. These operators are related to the creation and annihilation operators for the string oscillators, which describe the excitations of the string. In particular, α^-2 creates a state with two units of string oscillation energy, and α^-1α^-1 creates a state with one unit of string oscillation energy in each of two different directions. These states correspond to higher energy and higher momentum excitations of the string.

Now, why are these states considered representations of SO(D-1)? This has to do with how they transform under rotations in D dimensions. The state created by α^-2 can be thought of as a vector, because it has two indices (one for each oscillator) and transforms like a vector under rotations. Similarly, the state created by α^-1α^-1 can be thought of as a tensor, because it has two indices and transforms like a tensor under rotations. Therefore, these states form a complete representation of SO(D-1), as they include both vector and tensor states.

I hope this helps to clarify the solution to exercise 1.5 in Polchinski's book. Remember, it's always important to have a solid understanding of the basics before moving on to more advanced topics. Keep studying and asking questions, and you'll continue to build your knowledge and understanding of superstrings.