general relativity - Matrix derivative of a matrix with constraints

I am looking for a general method to obtain derivative rules of a constrained matrix with respect to its matrix elements.

In the case of a symmetric matrix $S_{ij}$ (with $S_{ij}=S_{ji}$), one way to do that is the following (see Variation of the metric with respect to the metric). We say that a variation of a matrix element $\delta S_{ij}$ is the same than that of $\delta S_{ji}$, and thus

$$\delta S_{ij}=\frac{\delta S_{ij}+\delta S_{ji}}{2}=\frac{\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}}{2}\delta S_{kl}=\mathcal S_{ij;kl}\delta S_{kl}.$$ The tensor $\mathcal S_{ij;kl}$ has the nice property that $\mathcal S_{ij;kl}\mathcal S_{kl;mn}=\mathcal S_{ij;mn}$. One then says that $$\frac{\delta S_{ij}}{\delta S_{kl}}=\mathcal S_{ij;kl}.$$

I must admit that why this is the correct procedure is not quite clear to me (that seems to be quite arbitrary, although obviously it works to compute derivatives of a function of a symmetric matrix). This means that it is not clear to me how to generalize that when the constraint is different.

For example, let's take the set of matrices $O$ belonging to the group $SO(N)$. Is there a way to write $\frac{\delta O_{ij}}{\delta O_{kl}}$ in terms of a tensor $\mathcal B_{ij;kl}$, with all the same nice properties ?

In the case of $SO(2)$, this seems quite easy, since then $O_{ji}=(-1)^{i+j}O_{ij}$, and one finds in that case

$$\frac{\delta O_{ij}}{\delta O_{kl}}=\frac{\delta_{ik}\delta_{jl}+(-1)^{i+j}\delta_{il}\delta_{jk}}{2},$$ which indeed does the job. Note however that I haven't use the defining property of $SO(N)$, that is $O O^T=1$, and I am not sure if this is relevant...

Already in the case of $SO(3)$, it does not seem to be easy to find the equivalent tensor...

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Side note : using the defining property of $SO(2)$, one can massage the formulas to obtain

$$\frac{\delta O_{ij}}{\delta O_{kl}}=-O_{il}O_{kj}.$$ First of all, it depends explicitly on $O$, which seems bad. Furthermore, if we tentatively define $\mathcal B_{ij;kl}[O]=-O_{il}O_{kj}$ (which already is different from what we found for $SO(2)$), then we have $\mathcal B_{ij;kl}[O]\mathcal B_{kl;mn}[O]=\delta_{im}\delta_{jn}$, which seems pretty weird...

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If anyone knows the standard procedure (if any exists) or a good reference, that would be greatly appreciated. In any case, a nice explanation (maybe a bit formal) in the case of symmetric matrix might also help me to get my head around the problem.

Answer

1. 
   

   Setup. Let there be given an $m$-dimensional manifold $M$ with coordinates $(x^1, \ldots, x^m)$. Let there be given an $n$-dimensional physical submanifold $N$ with physical coordinates $(y^1, \ldots, y^n)$. Let there be given $m-n$ independent constraints

   $$\chi^1(x)~\approx~ 0,\quad \ldots,\quad \chi^{m-n}(x)~\approx~ 0, \tag{1}$$ which defines the physical submanifold $N$. [Here the $\approx$ symbol means weak equality, i.e. equality modulo the constraints.] Assume that $$(y^1, \ldots, y^n, \chi^1, \ldots,\chi^{m-n})\tag{2}$$ constitutes a coordinate system for the extended manifold $M$.
   
   


2. 
   

   Dirac derivative. In analogy with the Dirac bracket, let us introduce a Dirac derivative

   $$\left(\frac{\partial}{\partial x^i}\right)_{\! D} ~:=~\frac{\partial}{\partial x^i} -\sum_{a=1}^{m-n}\frac{\partial \chi^a}{\partial x^i}\left(\frac{\partial }{\partial \chi^a}\right)_{\! y} ~=~\sum_{\alpha=1}^n\frac{\partial y^{\alpha}}{\partial x^i} \left(\frac{\partial }{\partial y^{\alpha}}\right)_{\! \chi} , \qquad i~\in~\{1,\ldots, m\}, \tag{3}$$ that projects onto the physical submanifold $$\left(\frac{\partial}{\partial x^i}\right)_{\! D} y^{\alpha}~=~\frac{\partial y^{\alpha}}{\partial x^i} , \qquad \left(\frac{\partial}{\partial x^i}\right)_{\! D} \chi^a~=~0,$$ $$i~\in~\{1,\ldots, m\},\qquad \alpha~\in~\{1,\ldots, n\},\qquad a~\in~\{1,\ldots, m\!-\!n\}.\tag{4}$$
   
   
   


3. 
   

   Remark. In many important cases it is possible to choose the physical coordinates $(y^1, \ldots, y^n)$ such that the Dirac derivative (4) can be written as linear combinations of unconstrained partial $x$-derivatives only, without referring to the $(y,\chi)$-coordinate system (2), cf. eqs. (10) & (14) below.

   
   


4. 
   

   Does Dirac derivatives commute? Does the commutator

   $$\left[ \left(\frac{\partial}{\partial x^i} \right)_{\! D}, \left(\frac{\partial}{\partial x^j}\right)_{\! D}\right] ~=~\sum_{\alpha,\beta=1}^n\frac{\partial y^{\alpha}}{\partial x^i}\left[\left(\frac{\partial }{\partial y^{\alpha}}\right)_{\! \chi}, \frac{\partial y^{\beta}}{\partial x^j}\right] \left(\frac{\partial }{\partial y^{\beta}}\right)_{\! \chi} - (i\leftrightarrow j)~\stackrel{?}{\approx}~0 \tag{5}$$ vanishes weakly? Not necessarily. But if the coordinate transformation $x^i \leftrightarrow (y^{\alpha}, \chi^a)$ is linear, then the Dirac derivatives commute.
   
   


5. 
   

   Example. Let the physical subspace be the hyperplane $N=\{\chi(x)=0\}$ with the constraint

   $$\chi~=~\sum_{i=1}^m x^i.\tag{6}$$ Define physical coordinates $$y^{\alpha}~=~ x^{\alpha} -\frac{1}{m}\sum_{i=1}^m x^i, \qquad \alpha~\in~\{1,\ldots, n\!=\!m\!-\!1\}.\tag{7}$$ Conversely, $$x^{\alpha}~=~y^{\alpha} +\frac{1}{m}\chi, \qquad \alpha~\in~\{1,\ldots, n\}, \qquad x^m~=~-\sum_{\beta=1}^n y^{\beta}+\frac{1}{m}\chi.\tag{8}$$ The derivatives are related as $$\frac{\partial}{\partial x^{\alpha}} ~=~\left(\frac{\partial }{\partial y^{\alpha}}\right)_{\! \chi} -\frac{1}{m} \sum_{\beta=1}^n\left(\frac{\partial }{\partial y^{\beta}}\right)_{\! \chi} +\left(\frac{\partial }{\partial \chi}\right)_{\! y}, \qquad \alpha~\in~\{1,\ldots, n\},$$ $$\frac{\partial}{\partial x^m} ~=~ -\frac{1}{m} \sum_{\beta=1}^n\left(\frac{\partial }{\partial y^{\beta}}\right)_{\! \chi} +\left(\frac{\partial }{\partial \chi}\right)_{\! y}.\tag{9}$$ The Dirac derivative becomes after some algebra $$\left(\frac{\partial}{\partial x^i}\right)_{\! D} ~=~\frac{\partial}{\partial x^i} -\left(\frac{\partial }{\partial \chi}\right)_{\! y} ~=~\frac{\partial}{\partial x^i} -\frac{1}{m} \sum_{j=1}^m\frac{\partial}{\partial x^j},\qquad i~\in~\{1,\ldots, m\}.\tag{10}$$
   
   



6. 
   

   Example. Differentiation wrt. a symmetric matrix can be viewed as a Dirac differentiation (3), where the constraints (1) are given by antisymmetric matrices. Define

   $$s_{(ij)}~:=~ \frac{M_{ij}+M_{ji}}{2} \quad\text{and}\quad a_{(ij)}~:=~ \frac{M_{ij}-M_{ji}}{2}\quad\text{for}\quad i~>~j; \quad\text{and}\quad d_{(i)}~:=~M_{ii}.\tag{11}$$ Conversely, $$M_{ij}~=~\theta_{ij}(s_{(ij)}+a_{(ij)})+\theta_{ji}(s_{(ji)}-a_{(ji)}) + \delta_{ij}d_{(i)} ,\tag{12}$$ where the discrete Heaviside step function $\theta_{ij}$ here is assumed to obey $\theta_{ii}=0$ (no implicit sum). The derivatives are related as $$\frac{\partial}{\partial M_{ij}} ~=~\frac{\theta_{ij}}{2}\left(\frac{\partial}{\partial s_{(ij)}}+\frac{\partial}{\partial a_{(ij)}}\right)+\frac{\theta_{ji}}{2}\left(\frac{\partial}{\partial s_{(ji)}}-\frac{\partial}{\partial a_{(ji)}}\right) + \delta_{ij}\frac{\partial}{\partial d_{(i)}} .\tag{13}$$ The Dirac derivative becomes after some algebra $$\left(\frac{\partial}{\partial M_{ij}}\right)_{\! D}~=~\frac{\theta_{ij}}{2}\frac{\partial}{\partial s_{(ij)}}+\frac{\theta_{ji}}{2}\frac{\partial}{\partial s_{(ji)}} + \delta_{ij}\frac{\partial}{\partial d_{(i)}} ~=~\frac{1}{2} \left(\frac{\partial}{\partial M_{ij}} +\frac{\partial}{\partial M_{ji}}\right).\tag{14}$$
   
   


7. 
   

   Remark. Additional complications arise if the coordinates and/or constraints are not globally defined. For starters, it is actually enough if (2) is a coordinate system in a tubular neighborhood of $N$.

   
   


8. 
   

   Reparametrizations of the constraints. Assume that there exists a second coordinate system

   $$(\tilde{y}^1, \ldots, \tilde{y}^n, \tilde{\chi}^1, \ldots,\tilde{\chi}^{m-n})\tag{15}$$ (which we adorn with tildes), such that $$\tilde{y}^{\alpha}~=~f^{\alpha}(y), \qquad \tilde{\chi}^a ~=~ g^a(y,\chi)~\approx~0.\tag{16}$$ This implies that $$\left(\frac{\partial }{\partial \chi^a}\right)_{\! y} ~=~ \left(\frac{\partial \tilde{\chi}^b}{\partial \chi^a}\right)_{\! y} \left(\frac{\partial }{\partial \tilde{\chi}^b}\right)_{\! \tilde{y}}, \qquad \left(\frac{\partial }{\partial y^{\alpha}}\right)_{\! \chi} ~\approx~\left(\frac{\partial \tilde{y}^{\beta}}{\partial y^{\alpha}}\right)_{\! \chi} \left(\frac{\partial }{\partial \tilde{y}^{\beta}}\right)_{\! \tilde{\chi}}, \tag{17}$$ i.e $$\Delta_{\chi}~:=~{\rm span}\left\{\left(\frac{\partial }{\partial \chi^1}\right)_{\! y}, \ldots ,\left(\frac{\partial }{\partial \chi^{n-m}}\right)_{\! y}\right\} ~\subseteq~ TM\tag{18}$$ is an involutive distribution, while $$\Delta_y~:=~{\rm span}\left\{\left(\frac{\partial }{\partial y^1}\right)_{\! \chi}, \ldots ,\left(\frac{\partial }{\partial y^n}\right)_{\! \chi}\right\} ~\subseteq~ TM\tag{19}$$ is a weak distribution.
   
   

   One may show that the Dirac derivative and its commutators

   $$\left(\frac{\partial}{\partial x^i}\right)^{\sim}_{\! D}~\approx~\left(\frac{\partial}{\partial x^i}\right)_{\! D}, \qquad \left[ \left(\frac{\partial}{\partial x^i} \right)^{\sim}_{\! D}, \left(\frac{\partial}{\partial x^j}\right)^{\sim}_{\! D}\right]~\approx~ \left[ \left(\frac{\partial}{\partial x^i} \right)_{\! D}, \left(\frac{\partial}{\partial x^j}\right)_{\! D}\right], \tag{20}$$ [wrt. the tilde and the untilde coordinate systems (15) and (2), respectively] agree weakly. This shows that the Dirac derivative (3) is a geometric construction.
   
   



9. 
   

   Subsubmanifold. Given a $p$-dimensional physical subsubmanifold $P$ with physical coordinates $(z^1,\ldots,z^p)$. Let there be given $n-p$ independent constraints

   $$\phi^1(y)~\approx~ 0,\quad \ldots,\quad \phi^{n-p}(y)~\approx~ 0, \tag{21}$$ which defines the physical submanifold $P$. Assume that $$(z^1, \ldots, z^p, \phi^1, \ldots,\phi^{n-p})\tag{22}$$ constitutes a coordinate system for the submanifold $N$. One may show that $$\left(\frac{\partial}{\partial x^i}\right)^{\!(P)}_{\! D} ~=~\left(\frac{\partial}{\partial x^i}\right)^{\!(N)}_{\! D} - \sum_{a=1}^{n-p}\left(\frac{\partial \phi^a}{\partial x^i}\right)^{\!(N)}_{\! D}\left(\frac{\partial }{\partial \phi^a}\right)_{\!z}, \qquad i~\in~\{1,\ldots, m\}. \tag{23}$$ This shows that the Dirac derivative construction behaves naturally wrt. further constraints.