Wave function in tensor product of Hilbert spaces If I had the wave function
$$\Psi\equiv\psi(r,\theta,\phi)\otimes\chi \in \mathscr{L}^2(\mathbb{R}^3)\otimes\mathbb{C}^{2S+1},$$
where $S$ is the spin of the state, is it correct to normalize the spin part of $\Psi$, namely $\chi$, regarding the spatial parameters $(r,\theta,\phi)$ as if they were fixed?
I mean: if $\psi\propto\sum Y_l^m (\theta,\phi)$, is it correct to say that the $Y_l^m(\theta,\phi)$'s are just numbers in $\mathbb{C}^{2S+1}$?
 A: The norm squared of your wavefunction is
$$\left<\Psi\big|\Psi\right>=\left<\psi\big|\psi\right>\left<\chi\big|\chi\right>$$
and this should be $1$. In particular, you can normalize both $\psi$ and $\chi$ separately and your $\Psi$ will be also normalized.
A: Actually, it seems you are asking a question about the definition of Hilbert space. A Hilbert space, as a vector space, is an inner product space -- that's the defining property of it. Therefore to define a Hilbert space, you have to define the inner product $\langle\psi_1|\psi_2\rangle$ of two vectors inside. 
The inner product, by definition, should be a scalar. Below we restrict our discussion on wave-functions in quantum physics. 
Firstly, for a simple Hilbert space structure $\mathcal{H}$, say, either $\psi$-space or $\chi$-space, when talking about wave-functions, it would be enough to define the scalar as a number. Therefore, when normalize vectors, i.e. wave-functions in $\mathcal{H}$, it is equivalent to require the result to be a number "$1$".
Now you are dealing with a tensor product space structure: $\mathcal{H}_0=\mathcal{H}_1\otimes\mathcal{H}_2$. The "vector" in $\mathcal{H}_0$ actually is defined as a tensor product of two simple vectors $\vec{v}_0 =\vec{v}_1\otimes\vec{v}_2$. Then the problem is what is the "scalar" in this $\mathcal{H}_0$? A more reasonable way to define a scalar is actually $a\otimes b$, where both $a, b\in\mathbb{C}$ are "daily-life" numbers. Therefore, more precisely, the normalization requirement now is modified as $\langle\vec{v}_0|\vec{v}_0\rangle = 1\otimes1\equiv \mathbb{1}$. That's why when you normalize it, you should do $\psi$ part and $\chi$ part separately.
Now you may ask: why don't I still use the old definition for scalar, as in a simple $\mathcal{H}$, i.e. just a "number"? 
Formally (mathematically), definition of tensor product space should not include any operation mixing them, therefore it's not a good idea to mix two "$1$" in the $1\otimes1$ structure -- for sure you could define another operation if you want, but it has nothing to do with inner product and normalization procedure.
Practically (physically), you could consider the following condition: I fix the spin part as $\chi = |1\rangle$ by turning on some external coupling acting only on the spin, and then vary the potential energy affecting spatial part. In this case, apparently the normalization of spatial part should not affect spin part at all -- which has already been fixed by hand.
