Spectrum of Operator on $L^2(X\times Y)$ I know that $L^2(X\times Y)$ is isomorphic to $L^2(X)\otimes L^2(Y)$. The proof is by defining bilienar form on $L^2(X) \times L^2(Y)$ and then by the universal property of tensor product. What will be the inverse of this map? Actually I have a description of an operator $u \in B(L^2(X\times Y))$. how can we write it interms of an operator $u' \in L^2(X)\otimes L^2(Y)$?
 A: The isomorphism is defined by the universal property of the tensor product. So we take $(f,g)\in L^2(X)\times L^2(Y)$ and map this to the function $f\otimes g$ on $X\times Y$ which is defined via $f\otimes g(x,y)=f(x)g(y)$. Now this bilinear map induces a linear map on the algebraic tensor product $L^2(X)\odot L^2(Y)\to L^2(X\times Y)$ and one checks that this preserves the inner product hence extends to $L^2(X)\otimes L^2(Y)$ and the extension is surjective, i.e. the Hilbert spaces $L^2(X\times Y)$ and $L^2(X)\otimes L^2(Y)$ are isomorphic.
You are now asking what is the inverse of the isomorphism. Well, take any function $h\in L^2(X\times Y)$. By the above proof, one sees that it can be approximated by a sequence of functions of the form $\bigg\{\sum_{j=1}^{N_k}f_{j,k}\otimes g_{j,k}\bigg\}_{k=1}^\infty$ where $f_{j,k}\in L^2(X)$ and $g_{j,k}\in L^2(Y)$. The inverse of $h$ will be the (unique and existent in $L^2(X)\otimes L^2(Y)$) limit of the functions $\sum_{j=1}^{N_k}f_{j,k}\odot g_{j,k}$ as $k\to\infty$. That's as descriptive as one can be in general, unless your function $h$ is itself of the form $\sum_{j=1}^Nf_j\otimes g_j$ in which case the inverse is simply $\sum_{j=1}^Nf_j\odot g_j$. Note that this operator is unitary (i.e. it is surjective and preserves inner product), so the inverse is equal to the adjoint.
The above isomorphism of Hilbert spaces, let's call it $\Phi:L^2(X)\otimes L^2(Y)\to L^2(X\times Y)$ induces a $*$-isomorphism $\Psi:\mathbb{B}(L^2(X)\otimes L^2(Y))\to\mathbb{B}(L^2(X\times Y))$ which is given by
$$\Psi(v)=\Phi\circ v\circ\Phi^* $$
So given an operator $u\in \mathbb{B}(L^2(X\times Y)$, you can describe it as an operator in $\mathbb{B}(L^2(X)\otimes L^2(Y))$ by taking $\Psi^{-1}(u)$, i.e. $u'=\Phi^*\circ u\circ \Phi$. This might seem that it is not very helpful, but you actually have a description of this operator on the linear span of the elementary tensors, which is a dense subspace.
I'm afraid that given the generality of your question one cannot be more specific than the above.
