AdS₃ Action in the Dreibein Formalism

We examine how to rewrite the (2 + 1)-dimensional AdS₃ action with gravitational field into action with dreibein. We obtained the non-trivial Einstein equation with dreibein. While Brown-Henneaux rewrite the AdS₃ action by using i) torsionless condition, ii) Einstein equation with dreibein. But we cannot use torsionless condition because the torsionless condition is automatically satisfied by imposing the dreibein hypothesis. Furthermore, we cannot use the Einstein equation with the dreibein to rewrite the AdS₃ action. If the AdS₃ action becomes in the Chern-Simons form, the equation of motion becomes trivial and there is no equation of motion, which contradicts with the non-trivial Einstein equation with dreibein. Therefore, the Brown-Henneaux’s procedure is not allowed, and we cannot obtain the Chern-Simons type action.

Introduction

In our previous papers [1]–[4], we have examined various problems concerning about the information loss from the black hole through the Hawking effect [5].

In connection with the holographic principle of ’t Hooft [6], Brown-Henneaux [7] claims that the (2 + 1)-dimensional AdS₃ theory with gravitational field corresponds to the 2-dimensional conformal field theory ${\text{CFT}}_2$ with dreibein. By rewriting the AdS₃ action with gravitational field into the action with dreibein, Brown-Henneaux claims that the resulting action with dreibein becomes that of the Chern-Simon type, so that the action becomes the surface term in the (2 + 1)-dimension theory, which becomes the 2-dimensional conformal field theory ${\text{CFT}}_2$. If we use the Brown-Henneaux’s result, the equation of motion in (2 + 1)-dimesnional Chern-Simons theory becomes trivial and there is no equation of motion. Then (2 + 1)-dimensional AdS₃ theory with gravitational field is considered to be the integrable theory as the resulting action with dreibein becomes the Chern-Simon type. Hence such mechanism is called ${\text{AdS}}_3/{\text{CFT}}_2$ correspondence. This is the prototype of Maldasena’s (3 + 1)-dimensional ${\text{AdS}}_5/{\text{CFT}}_4$ correspondence [8] in our world.

Brown-Henneaux’s claim is somewhat strange, because (2 + 1)-dimensional AdS₃ theory with gravitational field gives the non-trivial Einstein equation, and the solution of that Einstein equation gives the non-trivial exact solution [9]. While, if the action becomes the Chern-Simons type by rewriting the action with dreibein, the equation of motion becomes trivial and there is no equation of motion. Then we have some paradox. In our previous papers [1], [2], we have implicitly given resolution of this paradox.

In this paper, we pointed out that there are some problems for the Browwn-Henneaux’s procedure to rewrite AdS₃ action with gravitational fields into the action with dreibein. According to the Utiyama’s gauge theory of gravity [10], we take the orthodox procedure to rewrite the AdS₃ action with gravitational fields into the action with dreibein and we have found that the paradox does not appear.

Utiyama’s Gauge Theory of Gravity

In order that ${\mathit{\text{g}}}_{\mu \nu }$ becomes the gravitational field, we must put the condition of the parallel transport for ${\mathit{\text{g}}}_{\mu \nu }$ in the form

(1) $$\begin{array}{r}{\mathrm{\nabla }}_{\alpha }{\mathit{\text{g}}}_{\mu \nu }={\mathrm{\partial }}_{\alpha }{\mathit{\text{g}}}_{\mu \nu }-{\mathrm{\Gamma }}_{\mu \alpha }^{\lambda }{\mathit{\text{g}}}_{\lambda \nu }-{\mathrm{\Gamma }}_{\nu \alpha }^{\lambda }{\mathit{\text{g}}}_{\lambda \mu }=0.\end{array}$$

By using the condition ${\mathrm{\Gamma }}_{\mu \nu }^{\lambda }={\mathrm{\Gamma }}_{\nu \mu }^{\lambda }$, we must strongly solve ${\mathrm{\Gamma }}_{\mu \nu }^{\alpha }$ as the function of ${\mathit{\text{g}}}_{\mu \nu }$ and its derivative in the form

(2) $$\begin{array}{r}{\mathrm{\Gamma }}_{\mu \nu }^{\lambda }={\mathit{\text{g}}}^{\lambda \mathrm{\tau }}{\mathrm{\Gamma }}_{\mathrm{\tau }\mu \nu }=\frac{1}{2}{\mathit{\text{g}}}^{\lambda \sigma }({\mathrm{\partial }}_{\mu }{\mathit{\text{g}}}_{\sigma \nu }+{\mathrm{\partial }}_{\nu }{\mathit{\text{g}}}_{\sigma \mu }-{\mathrm{\partial }}_{\sigma }{\mathit{\text{g}}}_{\mu \nu }).\end{array}$$

There are various fermionic fields in our world, so that there are various Dirac equations for various fermionic fields. Then, in order to make the system which contain various fermionic fields to be consistent with the theory of general relativity, we must introduce vierbein fields $e_{\mu }^a=e^{a\tau }{\mathit{\text{g}}}_{\tau \mu }$, which satisfies ${\mathit{\text{g}}}_{\mu \nu }=e_{\mu }^a{e}_{a\nu }$.

In Utiyama’s gauge theory of gravity, by using the spin connection $A_{\mu }^{ab}=-A_{\mu }^{ba}=A^{ab\tau }{\mathit{\text{g}}}_{\tau \mu }$ as the gauge field, we define the covariant derivative of $e_{\mu }^a$ in the form

(3) $$\begin{array}{r}{D}_{\mu }{e}_{\nu }^a={\mathrm{\partial }}_{\mu }{e}_{\nu }^a-{\mathrm{\Gamma }}_{\mu \nu }^{\lambda }{e}_{\lambda }^a+A_{\mu }^{ab}{e}_{b\nu }.\end{array}$$

By using the covariant derivative of $e_{\mu }^a$, we can express (1) in the form

(4) $$\begin{array}{r}{\mathrm{\nabla }}_{\alpha }{\mathit{\text{g}}}_{\mu \nu }=(D_{\alpha }{e}_{a\mu })e_{\nu }^a+e_{a\mu }(D_{\alpha }{e}_{\nu }^a)=0.\end{array}$$

One of various ways to satisfy (4) is to put the vierbein hypothesis, that is, to put the covariant derivative of the vierbein to be zero in the form

(5) $$\begin{array}{r}{D}_{\mu }{e}_{\nu }^a={\mathrm{\partial }}_{\mu }{e}_{\nu }^a-{\mathrm{\Gamma }}_{\mu \nu }^{\lambda }{e}_{\lambda }^a+A_{\mu }^{ab}{e}_{b\nu }=0.\end{array}$$

By using this vierbein hypothesis, we must strongly solve $A_{\mu }^{ab}$ as the function of $e_{\mu }^a$ and its derivative. In the standard gauge theory, the gauge field is introduced as the independent field. While, as Utiyama theory is the gauge theory of gravity, $A_{\mu }^{ab}$ is not the independent field and $A_{\mu }^{ab}$ is expressed as the function of $e_{\mu }^a$ and its drivative, just as ${\mathrm{\Gamma }}_{\mu \nu }^{\lambda }$ is not the independent field and ${\mathrm{\Gamma }}_{\mu \nu }^{\lambda }$ is expressed as the function of ${\mathit{\text{g}}}_{\mu \nu }$ and its drivative.

We rewrite (5) in the form

(6) $$\begin{array}{r}{\mathrm{\Gamma }}_{\rho \mu \nu }=e_{a\rho }{\mathrm{\partial }}_{\mu }{e}_{\nu }^a+A_{\mu }^{ab}{e}_{b\nu }{e}_{a\rho }\end{array}$$

While, using we obtain

(7) $$\begin{array}{r}{e}_{a\rho }{\mathrm{\partial }}_{\mu }{e}_{\nu }^a={\mathrm{\partial }}_{\mu }(e_{a\rho }{e}_{\nu }^a)-e_{\nu }^a{\mathrm{\partial }}_{\mu }{e}_{a\rho }=\frac{1}{2}{\mathrm{\partial }}_{\mu }{\mathit{\text{g}}}_{\rho \nu }+\frac{1}{2}(e_{a\rho }{\mathrm{\partial }}_{\mu }{e}_{\nu }^a-e_{\nu }^a{\mathrm{\partial }}_{\mu }{e}_{a\rho }),\end{array}$$

(8) $$\begin{array}{rl}{\mathrm{\Gamma }}_{\rho \mu \nu }& =\frac{1}{2}{\mathrm{\partial }}_{\mu }{\mathit{\text{g}}}_{\rho \nu }+\frac{1}{2}(e_{a\rho }{\mathrm{\partial }}_{\mu }{e}_{\nu }^a-e_{\nu }^a{\mathrm{\partial }}_{\mu }{e}_{a\rho })+\frac{1}{2}{A}_{\mu }^{ab}(e_{a\rho }{e}_{b\nu }-e_{a\nu }{e}_{b\rho })\\ & =\frac{1}{2}({\mathrm{\partial }}_{\nu }{\mathit{\text{g}}}_{\rho \mu }+{\mathrm{\partial }}_{\mu }{\mathit{\text{g}}}_{\rho \nu }-{\mathrm{\partial }}_{\rho }{\mathit{\text{g}}}_{\nu \mu }).\end{array}$$

Then we have

(9) $$\begin{array}{rl}{A}_{\mu }^{ab}{e}_{a\rho }{e}_{b\nu }& =\frac{1}{2}{A}_{\mu }^{ab}(e_{a\rho }{e}_{b\nu }-e_{a\nu }{e}_{b\rho })\\ & =\frac{1}{2}({\mathrm{\partial }}_{\nu }{\mathit{\text{g}}}_{\rho \mu }-{\mathrm{\partial }}_{\rho }{\mathit{\text{g}}}_{\nu \mu })-\frac{1}{2}(e_{a\rho }{\mathrm{\partial }}_{\mu }{e}_{\nu }^a-e_{b\nu }{\mathrm{\partial }}_{\mu }{e}_{\rho }^b).\end{array}$$

Hence we finally obtain where we substitute ${\mathit{\text{g}}}_{\mu \nu }=e_{\mu }^a{e}_{a\nu }$.

(10) $$\begin{array}{rcl}& & A_{\mu }^{ab}=-\frac{1}{2}{e}^{a\rho }{e}^{b\sigma }({\mathrm{\partial }}_{\rho }{\mathit{\text{g}}}_{\sigma \mu }-{\mathrm{\partial }}_{\sigma }{\mathit{\text{g}}}_{\rho \mu })+\frac{1}{2}(e^{a\tau }{\mathrm{\partial }}_{\mu }{e}_{\tau }^b-e^{b\tau }{\mathrm{\partial }}_{\mu }{e}_{\tau }^a),\end{array}$$

Next, we relate the field strength of the gauge field $F_{\mu \nu }^{ab}={\mathit{\text{g}}}_{\mu \rho }{\mathit{\text{g}}}_{\nu \sigma }{F}^{ab\rho \sigma }$ and the Riemann tensor $R_{\beta \mu \nu }^{\alpha }={\mathit{\text{g}}}^{\alpha \tau }{R}_{\tau \beta \mu \nu }$. By using the identity ${\mathrm{\partial }}_{\nu }({\mathrm{\partial }}_{\mu }{e}_{\lambda }^a)-{\mathrm{\partial }}_{\mu }({\mathrm{\partial }}_{\nu }{e}_{\lambda }^a)=0$ and (5), we obtain

(11) $$\begin{array}{rl}0& ={\mathrm{\partial }}_{\nu }({\mathrm{\Gamma }}_{\mu \lambda }^{\tau }{e}_{\tau }^a)-{\mathrm{\partial }}_{\mu }({\mathrm{\Gamma }}_{\nu \lambda }^{\tau }{e}_{\tau }^a)-{\mathrm{\partial }}_{\nu }(A_{\mu }^{ab}{e}_{b\lambda })+{\mathrm{\partial }}_{\mu }(A_{\nu }^{ab}{e}_{b\lambda })\\ & =-R_{\lambda \mu \nu }^{\xi }{e}_{\xi }^a+F_{\mu \nu }^{ab}{e}_{b\lambda },\end{array}$$

(12) $$\begin{array}{rcl}& & R_{\lambda \mu \nu }^{\xi }={\mathrm{\partial }}_{\mu }{\mathrm{\Gamma }}_{\lambda \nu }^{\xi }-{\mathrm{\partial }}_{\nu }{\mathrm{\Gamma }}_{\lambda \mu }^{\xi }+{\mathrm{\Gamma }}_{\tau \mu }^{\xi }{\mathrm{\Gamma }}_{\lambda \nu }^{\tau }-{\mathrm{\Gamma }}_{\tau \nu }^{\xi }{\mathrm{\Gamma }}_{\lambda \mu }^{\tau },\end{array}$$

(13) $$\begin{array}{rcl}& & F_{\mu \nu }^{ab}={\mathrm{\partial }}_{\mu }{A}_{\nu }^{ab}-{\mathrm{\partial }}_{\nu }{A}_{\mu }^{ab}+A_{\mu }^{ac}{\eta }_{cd}{A}_{\nu }^{db}-A_{\nu }^{ac}{\eta }_{cd}{A}_{\mu }^{db}.\end{array}$$

Then we obtain

(14) $$\begin{array}{r}{R}_{\beta \mu \nu }^{\alpha }=e_a^{\alpha }{e}_{b\beta }{F}_{\mu \nu }^{ab}.\end{array}$$

By using (10), (14) is automatically satisfied. As there are many terms, it is quite messy to check the above identity (14) explicitly. But we can easily check that the second derivative terms of both side of (14) coincides in the following way. We start from the right-hand side of (14) in the form where we used (10). Therefore, the second derivative term of both side of (14) coincides.

(15) $$\begin{array}{ll}{e}_a^{\alpha }{e}_{b\beta }{F}_{\mu \nu }^{ab}& =e_a^{\alpha }{e}_{b\beta }({\mathrm{\partial }}_{\mu }{A}_{\nu }^{ab}-{\mathrm{\partial }}_{\nu }{A}_{\mu }^{ab}+A_{\mu }^{ac}{\eta }_{cd}{A}_{\nu }^{db}-A_{\nu }^{ac}{\eta }_{cd}{A}_{\mu }^{db})\\ & ={\mathrm{\partial }}_{\mu }(e_a^{\alpha }{e}_{b\beta }{A}_{\nu }^{ab})-{\mathrm{\partial }}_{\nu }(e_a^{\alpha }{e}_{b\beta }{A}_{\mu }^{ab})+\mathrm{????}(\text{products of the first derivative})\\ & ={\mathrm{\partial }}_{\mu }\left[e_a^{\alpha }{e}_{b\beta }\{-\frac{1}{2}{e}^{a\tau }{e}^{b\xi }({\mathrm{\partial }}_{\tau }{\mathit{\text{g}}}_{\xi \nu }-{\mathrm{\partial }}_{\xi }{\mathit{\text{g}}}_{\tau \nu })+\frac{1}{2}(e^{a\tau }{\mathrm{\partial }}_{\nu }{e}_{\tau }^b-e^{b\tau }{\mathrm{\partial }}_{\nu }{e}_{\tau }^a)\}\right]\\ & -(\mu \leftrightarrow \nu )+\mathrm{????}(\text{products of the first derivative})\\ & =\frac{1}{2}{\mathrm{\partial }}_{\mu }[-{\mathit{\text{g}}}^{\alpha \tau }({\mathrm{\partial }}_{\tau }{\mathit{\text{g}}}_{\beta \nu }-{\mathrm{\partial }}_{\beta }{\mathit{\text{g}}}_{\tau \nu })+{\mathit{\text{g}}}^{\alpha \tau }{\mathrm{\partial }}_{\nu }{\mathit{\text{g}}}_{\beta \tau }-2e^{a\alpha }{\mathrm{\partial }}_{\nu }{e}_{a\beta }]\\ & -(\mu \leftrightarrow \nu )+\mathrm{????}(\text{products of the first derivative})\\ & ={\mathrm{\partial }}_{\mu }{\mathrm{\Gamma }}_{\beta \nu }^{\alpha }-{\mathrm{\partial }}_{\nu }{\mathrm{\Gamma }}_{\beta \mu }^{\alpha }+\mathrm{????}(\text{products of the first derivative}),\end{array}$$

Using (14), we can rewrite the Lagrangian density in the form

(16) $$\begin{array}{r}\sqrt{-g}R=det(e)F_{\alpha \beta }^{ab}{e}_a^{\alpha }{e}_b^{\beta }.\end{array}$$

This is expressed in a more elegant way. By the definition of $det(e)$, we obtain

(17) $$\begin{array}{r}det(e){ϵ}^{abcd}={ϵ}^{\alpha \beta \mu \nu }{e}_{\alpha }^a{e}_{\beta }^b{e}_{\mu }^c{e}_{\nu }^d.\end{array}$$

Then we multiply ${ϵ}_{abef}{e}_c^{\rho }{e}_d^{\sigma }$ and take the sum, and we obtain

(18) $$\begin{array}{r}2det(e)(e_e^{\rho }{e}_f^{\sigma }-e_e^{\sigma }{e}_f^{\rho })={ϵ}^{\alpha \beta \rho \sigma }{ϵ}_{abef}{e}_{\alpha }^a{e}_{\beta }^b.\end{array}$$

Then (16) is expressed in the form where we substitute $g_{\mu \nu }=e_{\mu }^a{e}_{a\nu }$, (4).

(19) $$\begin{array}{r}\sqrt{-g}R=\frac{1}{2}det(e)F_{\alpha \beta }^{ab}(e_a^{\alpha }{e}_b^{\beta }-e_a^{\beta }{e}_b^{\alpha })=\frac{1}{4}{F}_{\alpha \beta }^{ab}{ϵ}_{abcd}{ϵ}^{\alpha \beta \mu \nu }{e}_{\mu }^c{e}_{\nu }^d,\end{array}$$

For the pure gravity case, there is another way to satisfy (4), that is, if we put $A_{\mu }^{ab}=0$ and substitute $g_{\mu \nu }=e_{\mu }^a{e}_{a\nu }$, (4) is satisfied. By using this shortcut way, we obtain the same Lagrangian density (19).

Rewriting AdS3 Action Into the Action with Dreibein

Here, we consider the AdS₃ theory in (2 + 1)-dimension. The action of the AdS₃ theory is given by

(20) $$\begin{array}{r}S=\frac{1}{16\pi G}\int d^{3}x\sqrt{-g}(R+\frac{2}{{\ell }^2}),\end{array}$$

Einstein equation with ${\mathit{\text{g}}}_{\mu \nu }$ is given by which gives

(21) $$\begin{array}{r}{R}_{\mu \nu }-\frac{1}{2}{\mathit{\text{g}}}_{\mu \nu }R=\frac{1}{{\ell }^2}{\mathit{\text{g}}}_{\mu \nu },\end{array}$$

(22) $$\begin{array}{r}{R}_{\mu \nu }=-\frac{2}{{\ell }^2}{\mathit{\text{g}}}_{\mu \nu },R=-\frac{6}{{\ell }^2}.\end{array}$$

By using the symmetry of indices, we obtain from the expression of $R_{\mu \nu }$. Combining (19) with (23), we obtain the Einstein equation with $e_{\mu }^a$ in the form

(23) $$\begin{array}{r}{R}_{\alpha \beta \mu \nu }=-\frac{1}{{\ell }^2}({\mathit{\text{g}}}_{\alpha \mu }{\mathit{\text{g}}}_{\beta \nu }-{\mathit{\text{g}}}_{\alpha \nu }{\mathit{\text{g}}}_{\beta \mu }),\end{array}$$

(24) $$\begin{array}{r}{R}_{\mu \nu }^{ab}=F_{\mu \nu }^{ab}=-\frac{1}{{\ell }^2}(e_{\mu }^a{e}_{\nu }^b-e_{\nu }^a{e}_{\mu }^b).\end{array}$$

Then we obtain

(25) $$\begin{array}{r}{\mathrm{\partial }}_{\mu }{A}_{\nu }^{ab}-{\mathrm{\partial }}_{\nu }{A}_{\mu }^{ab}+A_{\mu }^{ac}{\eta }_{cd}{A}_{\nu }^{db}-A_{\nu }^{ac}{\eta }_{cd}{A}_{\mu }^{db}+\frac{1}{{\ell }^2}(e_{\mu }^a{e}_{\nu }^b-e_{\nu }^a{e}_{\mu }^b)=0.\end{array}$$

While, from (5), the torsionless condition $D_{\mu }{e}_{\nu }^a-D_{\nu }{e}_{\mu }^a=0$ gives

(26) $$\begin{array}{r}{\mathrm{\partial }}_{\mu }{e}_{\nu }^a-{\mathrm{\partial }}_{\nu }{e}_{\mu }^a+A_{\mu }^{ab}{e}_{b\nu }-A_{\nu }^{ab}{e}_{b\mu }=0.\end{array}$$

This torsionless condition is necessary but not sufficient condition for the dreibein hypothesis (5), that is, (26) is automatically satisfied by using the dreibein hypothesis.

Brown-Henneaux solved ${\mathrm{\partial }}_{\mu }{e}_{\nu }^a-{\mathrm{\partial }}_{\nu }{e}_{\mu }^a$ and ${ϵ}^{abc}{A}_{\mu \nu }^{bc}$ by using (25) and (26). But we cannot use (26) as it is automatically satisfied. If we use the weaker condition (26), we obtain the wrong result. Therefore, the Einstein equation with $e_{\mu }^a$ is nothing but (25) itself. We must notice that we cannot use the Einstein equation to rewrite the action. From the Einstein equation, we obtain $R=-6/{\ell }^2$, but if we substitute this expression into the action, we obtain

which gives the wrong result, that is, we cannot obtain the Einstein (21). Brown-Henneaux solved (25) and (26) for ${\mathrm{\partial }}_{\mu }{e}_{\nu }^a-{\mathrm{\partial }}_{\nu }{e}_{\mu }^a$ and ${ϵ}^{abc}{A}_{\mu \nu }^{bc}$ and further substitute the solved expression into the action. Such procedure is not allowed in the following two reasons; i) we cannot use (26) as it is identically satisfied, ii) we cannot substitute solved expression into the action because we cannot use the Einstein equation to rewrite the action.

(27) $$\begin{array}{r}S=\frac{1}{16\pi G}\int d^{3}x(R+\frac{2}{{\ell }^2})\to \frac{1}{16\pi G}\int d^{3}x(-\frac{4}{{\ell }^2}),\end{array}$$

From the result of Utiyama’s theory, we obtain the action expressed with the dreibein in the form

where we substitute ${\mathit{\text{g}}}_{\mu \nu }=e_{\mu }^a{e}_{a\nu }$.

(28) $$\begin{array}{r}S=\frac{1}{16\pi G}\int d^{3}x\sqrt{-\mathit{\text{g}}}(R+\frac{2}{{\ell }^2})=\frac{1}{16\pi G}\int d^{3}xdet(e)(F_{\mu \nu }^{ab}{e}_a^{\mu }{e}_b^{\nu }+\frac{2}{{\ell }^2}),\end{array}$$

Summary and Discussions

We examine how to express the (2 + 1)-dimensional AdS₃ action with ${\mathit{\text{g}}}_{\mu \nu }$ into the action with $e_{\mu }^a$. We obtained the non-trivial Einstein equation with $e_{\mu }^a$. While Brown-Henneaux rewrite the AdS₃ action with ${\mathit{\text{g}}}_{\mu \nu }$ by using i) torsionless condition, ii) Einstein equation with $e_{\mu }^a$. But we cannot use torsionless condition because the torsionless condition is automatically satisfied. Furthermore, we cannot use the Einstein equation with $e_{\mu }^a$ to rewrite the action. If the resulting action with $e_{\mu }^a$ becomes in the Chern-Simons form, the equation of motion with $e_{\mu }^a$ becomes trivial and there is no equation of motion, which contradicts with the non-trivial Einstein equation with $e_{\mu }^a$.

There are various fermion fields in our world, so that the introduction of vierbein is inevitable to make the whole system to be consistent with the general relativity. Therefore, the compositeness of gravitational field ${\mathit{\text{g}}}_{\mu \nu }=e_{\mu }^a{e}_{a\nu }$ is inevitable.

In the classical theory of the composite gravitational field, we have the following questions. Is it possible to dissolve ${\mathit{\text{g}}}_{\mu \nu }$ into two vierbeins $e_{\mu }^a$ and $e_{a\nu }$? Especially, why the vierbein wave does not appear instead of the gravitational wave?

For composite gravitational field, there is the difficulty to quantize the theory. We do not know how to quantize the theory of kinetic terms being composite fields. By using the Nambu-Jona-Lasinio model [11], we can treat the composite state, but the kinetic term is not the composite fields, so that there is no problem to quantize.

In the superstring theory, we can construct spin 2 state as the composite of the two fermionic string modes, and we consider such spin 2 composite state as the graviton. Then gravitational field is the quantized graviton. Then we want to ask the following question. What kind of dynamical mechanism to condensate graviton as the composite state of two fermionic string modes, such as the condensation of $⟨\overline{\psi }\psi ⟩$ in the Nambu-Jona-Lasinio theory. If there is no such mechanism to condensate, the composite graviton can easily dissolve into two fermionic string modes. In the superstring theory, as there are various fermionic fields, so that there are various Dirac equations. Then we must introduce the bosonic vielbein to make the superstring theory to be consistent with the general relativity. Then, as the constituent of the composite gravitational field, we obtain the fermionic string mode and simultaneously the bosonic vielbein, but it contradicts.