# Tangent space of smooth Hilbert submanifolds

Let $$X, Y$$ be Hilbert spaces and $$F:X \rightarrow Y$$ smooth. Assume that $$M := F^{-1}(0) \subset X$$ is a smooth submanifold. Is it true that for any $$x\in M$$, the tangent space $$T_xM$$ is a Hilbert subspace of $$\mathrm{ker} D_xF$$?

Of course, if $$0$$ is a regular value of $$F$$, then by the implicit function theorem, $$M\subset X$$ is a smooth submanifold and $$T_xM = \mathrm{ker} D_xF$$.

What if $$0$$ is not a regular value?

• I know next to nothing about infinite dimensional manifolds, but why is the answer not "of course"? What is the subtlety? Sep 28 '20 at 13:09
• indeed the answer is "of course", nothing changes wrto finite dimension, even for $X,Y$ Banach manifolds. Sep 28 '20 at 13:39

True: via a local chart we can assume $$F^{-1}(0)$$ is a closed linear subspace $$N$$ of $$X$$, and since $$F_{|N}=0$$, we also have $$N\subset \text{ker} DF(x)$$.
(A formal explanation of the latter: if we denote $$i_N:N\to X$$ the (bounded, linear) inclusion map, $$F_{|N}=F\circ i_N:N\to Y$$ is the null map and by the chain rule $$0=D(F_{|N})(x)= D(F\circ i_N)(x) = DF(i_N(x))\circ i_N=DF(x)_{|N}$$ that is $$N\subset \text{ker} DF(x)$$ ).