Bosonic Strings and their Verma modules

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SUMMARY

The discussion focuses on the relationship between bosonic string fields and their Verma modules, specifically addressing the conformal dimensions of states. It establishes that bosonic string fields $X^{\mu}$ possess zero conformal dimension ($h=0$), while the highest weight state in a Verma module for open strings is defined by $L_0 \vert h \rangle = 1 \vert h \rangle$. The conversation highlights the equivalence of Verma modules with $h=1$ and $h=0$ in the absence of central charge in the Virasoro algebra, emphasizing the role of lowering operators in constructing these states.

PREREQUISITES
  • Understanding of bosonic string theory and its fields
  • Familiarity with representation theory and Verma modules
  • Knowledge of Virasoro algebra and its central extensions
  • Proficiency in conformal field theory concepts
NEXT STEPS
  • Study the implications of central charge in Virasoro algebra
  • Explore the construction and properties of Verma modules in detail
  • Investigate the role of lowering operators in representation theory
  • Learn about the applications of bosonic strings in theoretical physics
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The discussion is beneficial for theoretical physicists, string theorists, and researchers in representation theory who are exploring the connections between bosonic strings and their mathematical frameworks.

Korybut
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Hi there!

I have som troubles with representation theory.
It is obvious that bosonic strings fields $X^{\mu}$ has zero conformal dimension $h=0$. But when one builds Verma module (open string for example) highest weight state has the following definition
$$
L_0 \vert h \rangle = 1 \vert h \rangle
$$

All descendants states have even higher grading with respect to $L_0$. This shift is necessary for the absence or spurious states (negative norm).

How Verma module with $h=1$ coonected to open bosonic string with $h=0$?
 
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If Virasoro algebra has not central charge, Verma modules with $h=1$ and $h=0$ are in some sense equivalent
$$
\vert 1 \rangle = L_+ \vert 0 \rangle,
$$
where
$$
L_0 \vert 0\rangle =0 \;\; L_0 \vert 1 \rangle=-\vert 1 \rangle
$$
Applying lowering operators $L_-$
$$
L_- L_+ \vert 0\rangle = (L_- L_+ - 2L_0) \vert 0 \rangle =0
$$
State $\vert 1 \rangle$ is highest weight state from this simple exercise. But when Virasoro algebra has non-trivial central extension state $\vert 0 \rangle$ will apper in spectrum
 
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