Euler Lagrange equation of a functional on the space of traceless symmetric matrices

Question

Let $\mathcal{S}_0:\{Q\in \mathbb{R}^{3\times 3}:\text{tr}Q=0 \hspace{5pt}\text{and}\hspace{5pt} Q_{ij}=Q_{ji} \hspace{5pt} \text{for any $i,j=1,2,3$}\}$ and $$ \widetilde{ \mathcal{E}}(Q) =\int_{\mathbb{R}^3} \dfrac{1}{2}|\nabla Q|^2 +\left(\dfrac{1}{2} |Q|^2 - \sqrt{6} \text{tr} (Q^3) + \dfrac{1}{2} |Q|^4\right)dx. $$ on the space $\mathcal{S}_0$. The Euler Lagrange equation should be $$ -\Delta Q = -Q+3\sqrt{6}(Q^2-\dfrac{1}{3}|Q|^2Id) -2|Q|^2Q.$$

The question is I do not why I cannot put $Q+\varepsilon R$ as a variation and compute the equation, for $R \in \mathcal{S}_0$, and what correct variation should I use? Thank you!

Answer 1

For what it's worth, another way is to use a Lagrange multiplier$^1$ to enforce the traceless condition $${\rm tr}(Q)~=~0.\tag{1}$$ Then the extended functional becomes

$$\begin{align} S[Q,\lambda] ~=~&\int_{\mathbb{R}^3}\! d^3x\left( \frac{1}{2}{\rm tr}( \nabla Q^t\cdot \nabla Q)+ V(|Q|^2) -\sqrt{6}{\rm tr}(Q^3)+\lambda {\rm tr}(Q)\right),\cr |Q|^2~:=~&{\rm tr}(Q^tQ), \cr V(s)~=~&\frac{1}{2}s(1+s),\cr s~\in~\mathbb{R}. \end{align}\tag{2}$$

The Euler-Lagrange (EL) equation reads

$$ 0~=~\frac{\delta S}{\delta Q}~=~-\Delta Q^t + 2V^{\prime}(|Q|^2) Q^t-3\sqrt{6}Q^2 +\lambda \mathbb{1}_{3\times 3}.\tag{3}$$

We can determine the Lagrange multiplier by taking the trace of eq. (3):

$$ 0 ~=~-3\sqrt{6}{\rm tr}(Q^2)+3\lambda.\tag{4} $$

Eqs. (3) & (4) lead to OP's sought-for equation.

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$^1$ A similar idea can in principle be used to impose that the matrix $Q=Q^t$ is symmetric.

Answer 2

To adapt to the papers provided, let me consider just some open set $\Omega \subset \mathbb{R}^3$. As the admissible set of functions, I would suggest something like $$ M:=\{Q \in C^{2}(\Omega, \mathcal{S}_0) \mid E(Q)< \infty \}. $$ On this note, you could also choose a Sobolev space here, but I don't want to check all the emebddings to ensure that the energy is finite in all the scenarios for now. Moreover, we would like to compute the variation in the much bigger ambient space $$ N:=\{Q \in C^{2}(\Omega, \mathbb{R}^{3 \times 3}) \mid E(Q)< \infty \} $$ first. Let us therefore consider a variation $R \in C_c^{\infty}(\Omega, \mathbb{R}^{3 \times 3})$.This is also useful to get rid of any possible boundary terms that will appear when you integrate by parts. I will use the inner product $$ \langle A,B \rangle= \mathrm{tr}(AB^T) $$ on the space of matrices.

Then you can compute for any $R \in C_c^{\infty}(\Omega, \mathbb{R}^{3 \times 3})$ at for a function $Q \in M \subset N$ $$ 0=\frac{d}{d\epsilon}E(Q+\epsilon R)|_{\epsilon=0}, $$ which is well-defined. Let us examine the variation of the term $\int_{\Omega} | \nabla Q|^2 dx$: $$ \int_{\Omega} \langle \nabla Q, \nabla R \rangle dx=-\int_{\Omega} \langle \Delta Q, R \rangle dx $$ Here we integrated by parts and since $R$ has compact support, no boundary term appears. The terms $\int_{\Omega} \frac{1}{2}(|Q|^2+|Q|^4)dx$ is straightforward.
The last term of interest is $\int_{\Omega} \sqrt{6}\mathrm{tr} (Q^3)dx$. We note that $$ \mathrm{tr}(Q^3)=\langle Q^2,Q \rangle, $$ which has the expansion \begin{align*} \langle (Q+\epsilon R)^2,Q+\epsilon R \rangle&=\langle Q^2,Q \rangle+ \epsilon \langle Q^2,R \rangle +2 \epsilon \langle QR, Q \rangle +O(\epsilon^2) \\ &=\langle Q^2,Q \rangle+ \epsilon \langle Q^2,R \rangle +2 \epsilon \langle Q^2, R \rangle +O(\epsilon^2) \\ &= \langle Q^2,Q \rangle+3 \epsilon \langle Q^2,R \rangle + O(\epsilon^2) \end{align*} So in terms of the variation, we would end up with $$ \int_{\Omega} 3\sqrt{6} \langle Q^2,R \rangle dx. $$ If I put all terms together now, we get the variational equation $$ 0=\int_{\Omega} \langle -\Delta Q + Q - 3 \sqrt{6}Q^2+2 |Q|^2Q, R \rangle dx. $$ Now, we can think of $f:=-\Delta Q + Q - 3 \sqrt{6}Q^2+2 |Q|^2Q$ as the $L^2$-gradient of $E$ at $Q$ in the space $N$. But we only need it to be the $L^2$-gradient in the space $M$ and hence we just use the projection $$ P:N \to M, Q \mapsto Q-\frac{1}{3}\mathrm{tr}(Q)\mathrm{Id} $$ and calculate $Pf$. Note that all the terms of $f$, except for $Q^2$ are already symmetric and traceless. The term $Q^2$ is symmetric, but not traceless, and hence $$ PQ^2=Q^2-\frac{1}{3}|Q|^2 \mathrm{Id} $$ and therefore, $$ 0=Pf=-\Delta Q + Q - 3 \sqrt{6}(Q^2-\frac{1}{3}|Q|^2)+2 |Q|^2Q $$ as desired.