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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?

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    $\begingroup$ I know next to nothing about infinite dimensional manifolds, but why is the answer not "of course"? What is the subtlety? $\endgroup$
    – Nik Weaver
    Commented Sep 28, 2020 at 13:09
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    $\begingroup$ indeed the answer is "of course", nothing changes wrto finite dimension, even for $X,Y$ Banach manifolds. $\endgroup$
    – Pietro Majer
    Commented Sep 28, 2020 at 13:39

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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) $ ).

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  • $\begingroup$ Thank you! Somehow, I really missed that. $\endgroup$
    – user160747
    Commented Sep 28, 2020 at 14:09

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