# What is the determinant of the Wheeler-DeWitt metric tensor constructed from spatial metrics in ADM formalism?

The Hamiltonian constraint of General relativity has the following form

\begin{align} \frac{(2\kappa)}{\sqrt{h}}\left(h_{ac} h_{bd} - \frac{1}{D-1} h_{ab} h_{cd} \right)p^{ab} p^{cd} - \frac{\sqrt{h}}{(2\kappa)}\left({}^{(D)}\mathscr{R} - 2 \Lambda\right), \end{align} where $$D$$ is the number of spatial dimensions. Let's define

$$G_{abcd} \equiv \frac{1}{\sqrt{h}} \left(h_{a(c} h_{d)b} - \frac{1}{D-1}h_{ab} h_{cd} \right).$$

How to define the determinant of $$G_{abcd}$$?

• Hi @Faber Bosch. Reference to definition? Which page? Commented Jun 25 at 14:38
• What definition of determinant are you using? That's typically reserved for square matrices (rank 2 tensors). You have a rank-4 tensor. I'm not aware of any definition that's compatible with your question. Commented Jun 25 at 14:52
• Check out 'hyperdeterminant'. en.wikipedia.org/wiki/Hyperdeterminant Commented Jun 25 at 15:00

1. The Wheeler-DeWitt metric \begin{align} G~=~&G_{IJ}(\mathrm{d}y\odot\mathrm{d}y)^I\odot(\mathrm{d}y\odot\mathrm{d}y)^J\cr ~=~&G_{i_1i_2,j_1j_2}(\mathrm{d}y^{i_1}\odot\mathrm{d}y^{i_2})\odot(\mathrm{d}y^{j_1}\odot\mathrm{d}y^{j_2})\tag{1} \end{align} is by definition a covariant (0,4) spatial tensor of a $$D=d+1$$ dimensional split spacetime $${\cal M}$$ with local coordinates $$(t,y^1,\ldots, y^d)$$. The components \begin{align} G_{IJ}~=~&\frac{1}{2}(\gamma_{i_1j_1}\gamma_{i_2j_2}+\gamma_{i_1j_2}\gamma_{i_2j_1})-\frac{1}{d-1}\gamma_I\gamma_J~=~G_{JI},\cr G^{IJ}~=~&\frac{1}{2}(\gamma^{i_1j_1}\gamma^{i_2j_2}+\gamma^{i_1j_2}\gamma^{i_2j_1})-\gamma^I\gamma^J~=~G^{JI}, \end{align} \tag{2} are built from the components $$\gamma_I$$ of a spatial metric $$\gamma~=~\gamma_I(\mathrm{d}y\odot\mathrm{d}y)^I~=~\gamma_{i_1i_2}\mathrm{d}y^{i_1}\odot\mathrm{d}y^{i_2}.\tag{3}$$ Here $$I=i_1i_2=i_2i_1$$ is a symmetric double index, i.e. the index $$I$$ can take $$\frac{d(d+1)}{2}$$ values, where $$i_1,i_2\in\{1,\ldots, d\}$$ run over spatial indices.
2. Due to the positive signature $$(+,\ldots,+)$$ we can diagonalize the spatial metric as a unit matrix $$\gamma_{i_1i_2}~=~(m^Tm)_{i_1i_2}~=~m^{k_1}{}_{i_1}m^{k_1}{}_{i_2}, \qquad \det(\gamma_{..})~=~\det(m)^2,\tag{4}$$ where the $$d\times d$$ matrix $$m~=~\begin{pmatrix} \lambda_1 && \cr &\ddots& \cr && \lambda_d \end{pmatrix} O \tag{5}$$ is a diagonal matrix times an orthogonal matrix. If we define the $$\frac{d(d+1)}{2}\times\frac{d(d+1)}{2}$$ matrix $$M^I{}_J~:=~\frac{1}{2}(m^{i_1}{}_{j_1}m^{i_2}{}_{j_2}+m^{i_1}{}_{j_2}m^{i_2}{}_{j_1}),\tag{6}$$ then $$G_{IL}~=~(M^T)_I{}^J G^{(\gamma=\delta)}_{JK}M^K{}_L. \tag{7}$$
3. The determinant $$\det (G_{IJ})~=~\det(M)^2\det(G^{(\gamma=\delta)}_{IJ}) ~=~\frac{\det(\gamma_{..})^{d+1}}{1-d}\tag{8}$$ is over a $$\frac{d(d+1)}{2}\times\frac{d(d+1)}{2}$$ square matrix $$G_{IJ}$$, cf. OP's question.
4. Sketched proof of eq. (8). The matrix $$G^{(\gamma=\delta)}_{IJ}$$ has eigenvalue $$\frac{1}{1-d}$$ with multiplicity 1 and all other eigenvalues are 1, so the determinant is $$\det(G^{(\gamma=\delta)}_{IJ})~=~\frac{1}{1-d}. \tag{9}$$ We can ignore the orthogonal matrix $$O$$ as it cannot change the determinant. Then $$m$$ is a diagonal matrix. We calculate: \begin{align}\det(M)^2~=~&\left(\prod_{1\leq i\leq j\leq d} \lambda_i\lambda_j\right)^2 ~=~\left(\prod_{1\leq i,j\leq d} \lambda_i\lambda_j\right) \prod_{1\leq k\leq d}\lambda_k^2\cr ~=~& \left(\prod_{1\leq k\leq d}\lambda_k\right)^{2(d+1)}~=~\det(m)^{2(d+1)}~=~ \det(\gamma_{..})^{d+1}.\end{align}\tag{10} $$\Box$$
• Thank you for this excellent answer! Just to clarify the notation $\det (\gamma_{..})$ is simply the determinant of the spatial metric, right? I am confused about the double dots. Commented Jun 29 at 19:41