The vertex operator associated with massless state is

$$V(k,\epsilon) = -\frac{2}{\alpha}\epsilon_{\mu\nu}(k)\bar{\partial}X^\mu(\bar{z})\partial X^\nu(z)e^{ik\cdot X(z,\bar{z})}$$

The polarization tensor can be decomposed into symmetric (Graviton), antisymmetric and Trace bit (Dilaton) Lust,theisen 16.9

$$\epsilon^{(h)}_{\mu\nu} = \epsilon^{(h)}_{\nu\mu},\qquad\epsilon^{(h)}_{\mu\nu} \eta^{\mu\nu} = k^{\mu}\epsilon^{(h)}_{\mu\nu} = 0, $$

$$\epsilon^{(B)}_{\mu\nu} = -\epsilon^{(B)}_{\nu\mu},\qquad k^{\mu}\epsilon^{(B)}_{\mu\nu} = 0, $$

$$\epsilon^{(D)}_{\mu\nu} = \frac{1}{\sqrt{d-2}}(\eta_{\mu\nu}-k_{\mu}\bar{k_{\nu}}-k_{\nu}\bar{k_{\mu}})$$

$\bar{k}$ is an arbitrary light like vector orthogonal to $k$. Can you please tell me why we took that particular form for the Dilaton?

  • 1
    $\begingroup$ Why the dilaton term has this particular form is nicely and concisely explained here. $\endgroup$
    – Dilaton
    Sep 8 '14 at 10:13

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