# A proof of the Ibragimov-Kara-Mahomed commutation relation

Let $$u_a(x),\,a=1,2,\ldots n$$ be a $$n$$-component field in Minkowski spacetime $$x^\mu,\,\mu=0,1,2,3$$ and let $$u_{a,\,\mu}=\frac{du_a}{dx^\mu}$$. Let us introduce two operators (we use Einstein summation convention that repeated indices are implicitly summed over) $$\hat G=\tau^\mu\,\frac{d}{dx^\mu}+\sigma_a\,\frac{\partial}{\partial u_a}+ \sigma_{a,\mu}\,\frac{\partial}{\partial u_{a,\,\mu}}$$ and $$\hat N^\mu=\tau^\mu+\sigma_a\,\frac{\delta}{\delta u_{a,\,\mu}}+ \sigma_{a,\nu}\,\frac{\delta}{\delta u_{a,\,\mu\nu}},$$ where $$\sigma_a=\xi_a-\tau^\mu\,u_{a,\,\mu},$$ $$\xi_a(x,u)$$ and $$\tau^\mu(x,u)$$ do not depend on field derivatives, and $$\frac{\delta}{\delta u_{a,\,\mu}}=\frac{\partial}{\partial u_{a,\,\mu}}- \frac{d}{dx^\nu}\frac{\partial}{\partial u_{a,\,\mu\nu}},$$ $$\frac{\delta}{\delta u_{a,\,\mu\nu}}=\frac{\partial}{\partial u_{a,\,\mu\nu}}- \frac{d}{dx^\alpha}\frac{\partial}{\partial u_{a,\,\mu\nu\alpha}}.$$ It is claimed in http://link.springer.com/article/10.1023%2FA%3A1008240112483 (Lie–Bäcklund and Noether Symmetries with Applications, by N.H. Ibragimov, A.H. Kara and F.M. Mahomed) that $$[\hat G+\tau^\nu_{,\,\nu},\hat N^\mu]=\tau^\mu_{,\,\nu}\,\hat N^\nu. \tag{1}$$ The authors write that "the relation is proved by straightforward, albeit tedious, computation".

Is there a simpler way to prove the commutation relation (1)? I'm interested to prove it only in the first jet space $$(x^\mu,u_a,u_{a,\,\nu})$$, so only the first few terms from the cited article are presented in the definitions of $$\hat G$$ and $$\hat N^\mu$$.

Now I have found a proof. In the first jet space, it is sufficient to take $$\hat N^\mu=\tau^\mu+\sigma_a\,\frac{\partial}{\partial u_{a,\,\mu}},$$ and the proof proceeds as follows. We have $$[\hat G+\tau^\nu_{,\,\nu},\,\hat N^\mu]=\hat G(\tau^\mu)+\hat G(\sigma_a)\, \frac{\partial}{\partial u_{a,\,\mu}}+\sigma_a\left [\hat G,\,\frac{\partial} {\partial u_{a,\,\mu}}\right]+\sigma_a\left [\tau^\nu_{,\,\nu},\, \frac{\partial}{\partial u_{a,\,\mu}}\right]. \tag{1}$$ But $$\left [\tau^\nu_{,\,\nu},\,\frac{\partial}{\partial u_{a,\,\mu}}\right]= -\frac{\partial \tau^\nu_{,\,\nu}}{\partial u_{a,\,\mu}}=-\frac{\partial \tau^\mu}{\partial u_a}, \tag{2}$$ because $$\tau^\nu_{,\,\nu}=\frac{\partial \tau^\nu}{\partial x^\nu}+u_{b,\,\nu}\, \frac{\partial \tau^\nu}{\partial u_b}. \tag{3}$$ On the other hand, as $\tau^\mu$ doesn't depend on field derivatives, $$\hat G(\tau^\mu)=\tau^\nu\,\tau^\mu_{,\,\nu}+\sigma_a\,\frac{\partial \tau^\mu}{\partial u_a}. \tag{4}$$ Further we have $$\hat G(\sigma_a)=\tau^\nu\,\sigma_{a,\,\nu}+\sigma_b\,\frac{\partial \sigma_a}{\partial u_b}+\sigma_{b,\,\nu}\,\frac{\partial \sigma_a} {\partial u_{b,\,\nu}}. \tag{5}$$ But $$\sigma_a=\xi(x,u)-\tau^\mu(x,u)\,u_{a,\,\mu},$$ and, therefore, $$\frac{\partial \sigma_a}{\partial u_{b,\,\nu}}=-\delta_a^b\,\tau^\nu. \tag{6}$$ Substituting this into (5), we get $$\hat G(\sigma_a)=\sigma_b\,\frac{\partial \sigma_a}{\partial u_b}. \tag{7}$$ It remains to calculate the commutator $$\left [\hat G,\,\frac{\partial} {\partial u_{a,\,\mu}}\right]= \left [\tau^\nu \,\frac{\partial }{\partial x^\nu}+\xi_b\,\frac{\partial } {\partial u_b}+\eta_{b\nu}\,\frac{\partial }{\partial u_{b,\,\nu}},\, \frac{\partial} {\partial u_{a,\,\mu}}\right]=-\frac{\partial \eta_{b\nu}} {\partial u_{a,\,\mu}}\,\frac{\partial }{\partial u_{b,\,\nu}}, \tag{8}$$ where we have used the fact that $\tau^\nu$ and $\xi_a$ do not depend on field derivatives. Using $$\eta_{b\nu}=\xi_{b,\,\nu}-\tau^\alpha_{,\,\nu}\,u_{b,\,\alpha},$$ along with $$\frac{\partial \xi_{b,\,\nu}}{\partial u_{a,\,\mu}}=\delta^\mu_\nu\, \frac{\partial \xi_b}{\partial u_a},\;\;\;\frac{\partial \tau^\alpha_{,\,\nu}}{\partial u_{a,\,\mu}}=\delta^\mu_\nu\, \frac{\partial \tau^\alpha}{\partial u_a}, \tag{9}$$ we get $$\frac{\partial \eta_{b\nu}}{\partial u_{a,\,\mu}}=\delta^\mu_\nu\,\left ( \frac{\partial \xi_b}{\partial u_a}-u_{b,\,\alpha}\,\frac{\partial \tau^\alpha}{\partial u_a}\right )-\delta_a^b\,\tau^\mu_{,\,\nu}= \delta^\mu_\nu\,\frac{\partial \sigma_b}{\partial u_a}-\delta_a^b\, \tau^\mu_{,\,\nu}. \tag{10}$$ Therefore $$\left [\hat G,\,\frac{\partial} {\partial u_{a,\,\mu}}\right]= \tau^\mu_{,\,\nu}\,\frac{\partial }{\partial u_{a,\,\nu}}- \frac{\partial \sigma_b}{\partial u_a}\,\frac{\partial } {\partial u_{b,\,\mu}}. \tag{11}$$ Now (2), (4), (7) and (11), in combination with (1), imply the desired result: $$[\hat G+\tau^\nu_{,\,\nu},\,\hat N^\mu]=\tau^\mu_{,\,\nu}\,\left (\tau^\nu+ \sigma_a\,\frac{\partial }{\partial u_{a,\,\nu}}\right )=\tau^\mu_{,\,\nu}\, \hat N^\nu.$$