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Consider an unbounded self-adjoint operator $A$ on a Hilbert space $\mathcal{H}$. Let $\mathcal{J} \subset \mathcal{H}$ a closed subspace reducing $A$, i.e. such that $P A \subset A P$ where $P$ denotes orthogonal projection onto $J$. Equivalently, $P \mathcal{D}(A) \subset \mathcal{D}(A)$ and $P A \psi = A P \psi$ for all $\psi \in \mathcal{D}(A)$. Then the restriction $A|\mathcal{J}$ is a densely defined operator on $\mathcal{J}$ with domain $\mathcal{D}(A) \cap \mathcal{J}$.

My question is this: is $A|\mathcal{J}$ again self-adjoint?

The reason I am interested in this question is the following: Take $\mathcal{H} = L^2(\mathbb{R}^{3N})$, and $\mathcal{J} = \Lambda L^2(\mathbb{R}^{3N})$, where the $\Lambda$ denotes the totally antisymmetric subspace, i.e. those functions $\psi(\vec{x_1},...,\vec{x_N})$ with the property that for any permutation $\sigma \in S^N$,

\begin{equation} \psi(\vec{x_{\sigma(1)}},...,\vec{x_{\sigma(N)}}) = sign(\sigma) \psi(\vec{x_1},...,\vec{x_N}) \end{equation}

Then $\mathcal{H}$ is the phase space of an atom consisting of $N$ electrons, with the nucleus fixed at the origin, and $\mathcal{J}$ is the phase space for the same system, but respecting the Pauli principle. My operator on $\mathcal{H}$ is the self-adjoint operator given by

\begin{equation} H^N = - \sum_{j=1}^{N} \Delta_j + \sum_{j = 1}^{N} V_{en}(x_j) + \sum_{i < j} V_{ee}(x_i - x_j) \end{equation}

where the $V_{ee}$ terms denote electron-electron repulsion, and $V_{en}$ electron-nucleus attraction.

In fact, in this case I know of a proof: since the Fourier transform maps antisymmetric functions to antisymmetric functions, one can first show that $H_0 = - \Delta$ is self-adjoint when restricted to $\mathcal{J}$. Then use the fact that the remaining terms are $H_0$-bounded with $H_0$-bound $0$, which remains true for the restriction.

I am aware of the related question at Selfadjoint operators. However, the case I am interested in is very different, in the sense that the restricted operator $A|\mathcal{J}$ is considered an operator on $\mathcal{J}$ rather than on the full Hilbert space $\mathcal{H}$. Indeed, considering $A|\mathcal{J}$ to be an operator on $\mathcal{H}$, it is in general (and certainly in my case) not even densely defined.

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    A symmetric operator $A$ is selfadjoint iff $(A\pm iI)$ are surjective. No restriction of $A$ can have that property. A restriction could be essentially selfadjoint, but only if its closure is $A$. – 2017-01-31
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    @TrialAndError Yes, but is this also true if we restrict the operator to the subspace $\mathcal{J}$? I think the statement is true at least in the above example with $-\Delta$. – 2017-01-31
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    If $A : \mathcal{D}(A)\subseteq H \rightarrow H$ is selfadjoint, then $A\pm iI$ are injective and surjective. If you restrict $A$ to a subspace $J \subset\mathcal{D}(A)$, then $(A\pm i I)J \ne H$, which means that the restriction of $A$ to $J$ cannot be selfadjoint, though it might be essentially self-adjoint, with a selfadjoint closure. – 2017-01-31
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    I see what you mean. However the condition on $\mathcal{J}$ is precisely such that $A$ can be restricted to an operator $A| \mathcal{J}: \mathcal{D}(A) \cap \mathcal{J} \rightarrow \mathcal{J}$. So it might still be true that $(A \pm i I) \mathcal{J} = \mathcal{J}$. Clearly, in general $A | \mathcal{J}$ as an operator on $\mathcal{H}$ wont make sense because it wont even be densely defined. That's why I was referring to the similar question I linked to. – 2017-02-01
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    $A\pm iI$ are surjective and injective for a selfadjoint $A$. So $D_1 \ne D_2$ implies $(A+iI)D_1 \ne (A+iI)D_2$. – 2017-02-01
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    I don't see what you are trying to do with $D_1$ and $D_2$. Would you mind adding some clarification? – 2017-02-01
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    These can be as general as sets. In your cases, subspaces. If $A+iI$ is injective and surjective on its domain $\mathcal{D}(A)$, then $(A+iI)\mathcal{D}(A) = (A+iI)(\mathcal{D}(A)\cap\mathcal{J})$ iff $J=\mathcal{D}(A)$. This is a function mapping issue. – 2017-02-01
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52891/discussion-between-max-and-trialanderror). – 2017-02-01
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    I don't have time for chat. – 2017-02-01
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    @TrialAndError: I don't agree with your objection either. As a really trivial example, suppose $A$ is the identity operator. Then its restriction to any closed subspace is again self-adjoint (and in particular $A \pm iI$ are surjective on that subspace). – 2017-02-01
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    I have a question, though. Usually when you say something is an operator on $J$ you mean its domain and image are both subsets of $J$. In this case that would only be the case if $J$ is $A$-invariant but I don't see that you are assuming that. – 2017-02-01
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    I changed my assumptions: initially $A$-invariance of $\mathcal{J}$ was the only thing I assumed. Now I assume something much stronger: that the projection to $\mathcal{J}$ (call it $P$) preserves the domain of $A$, and that $P$ and $A$ commute for all vectors in $\mathcal{D}(A)$. In particular, if $\psi \in \mathcal{D}(A) \cap \mathcal{J}$, then $A \psi = A P \psi = P A \psi \in \mathcal{J}$. Hence I can consider $A$ to be an operator on $\mathcal{J}$ with domain $\mathcal{D}(A) \cap \mathcal{J}$. – 2017-02-01

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The answer is yes.

Since $A|\mathcal{J}$ certainly remains symmetric, it suffices to show that $(A \pm i I)(\mathcal{D}(A) \cap \mathcal{J}) = \mathcal{J}$. But we know that $A \pm i I : \mathcal{D}(A) \rightarrow \mathcal{H}$ are surjective since $A$ is self-adjoint. Hence for $\psi \in \mathcal{J} \subset \mathcal{H}$ there exist elements $\phi_{\pm} \in \mathcal{D}(A)$ such that $(A \pm i I)\phi_{\pm} = \psi$. Now replacing $\phi_{\pm}$ by $P \phi_{\pm}$ where $P$ is the orthongonal projection to $\mathcal{J}$, we see that $P \phi_{\pm} \in \mathcal{D}(A) \cap \mathcal{J}$ and $(A \pm i I)P \phi_{\pm} = P(A \pm i I) \phi_{\pm} = P \psi = \psi$.

Hence $A | \mathcal{J}: \mathcal{D}(A) \cap \mathcal{J} \rightarrow \mathcal{J}$ is self-adjoint.