$\DeclareMathOperator{\div}{div}$ $\def\bu{\mathbf{u}}$ Let $D$ be the square $[0,1]^2$ and consider the following space: $$ V:=\{\bu: \bu\in H^2(D)^2, \div \bu=0, u|_{\partial D}=0 \}. $$
Introduce also $$ H:=\{\bu: \bu\in L^2(D)^2, \div \bu=0\}. $$
Then the Stokes operator is defined as an operator $V\to H$ as $$ A\bu:=-P_L\Delta\bu, $$ where $P_L$ is the projection on $H$.
I don't understand, why the Stokes operator is not equal to minus Laplacian? Indeed, if $\bu\in V$, then $\div \bu=0$ and thus $$ \div \Delta\bu=u^1_{xxx}+u^1_{yyx}+u^2_{xxy}+u^2_{yyy}=(u^1_x+u^2_y)_{xx}+(u^1_x+u^2_y)_{yy}=\Delta\div\bu=0. $$ In this case, $\Delta \bu\in H$ and thus the projection is not needed.
Question: what is wrong in my reasoning? Why the Stokes operator is not equal to Laplacian? Related question: Can one write explicitly the eigenfunctions of Stokes operator in this simple situation? At least can maybe one bound $\lambda_1$, the smallest eigenvalue, from below?