Do Tensor Product Properties Hold in Infinite Dimensional Hilbert Spaces?

LagrangeEuler
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Is this correct in infinite dimensional Hilbert spaces?
## (\hat{A}_1 \otimes \hat{A}_2)^{-1}=\hat{A}^{-1}_1 \otimes \hat{A}^{-1}_2 ##
## (\hat{A}_1 \otimes \hat{A}_2)^{\dagger}=\hat{A}^{\dagger}_1 \otimes \hat{A}^{\dagger}_2 ##
## (\hat{A}_1 +\hat{A}_2) \otimes \hat{A}_3=(\hat{A}_1 \otimes \hat{A}_2)+(\hat{A}_2 \otimes \hat{A}_3) ##
## \hat{A}_1 \otimes (\hat{A}_2+\hat{A}_3)=(\hat{A}_1 \otimes \hat{A}_2)+(\hat{A}_1 \otimes \hat{A}_3) ##
## \hat{1} \otimes \hat{1}=\hat{1} ##
## (\hat{A}_1 \otimes 0)=(0 \otimes \hat{A}_2)=0 ##
Can you tell me a book where I can see this properties. I found this only for operators which acts in finite dimensional Hilbert spaces.
 
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Your question belongs to the part of mathematics usually discussed within "Operators theory". But here is essentially the positive answer to your questions (with somewhat different notation) - provided appropriate care is being taken:

stratila58.jpg


Taken from "Lectures on von Neumann Algebras" by Serban Stratila and Laszlo Zsido, Abacus Press 1975. You can also find it in online Bratteli and Robinson book "Operator algebras and quantum statistical mechanics", Vol. 1, but there it is more complicated as tensor product of not just two but a family of Hilbert spaces is being considered.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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