Tensor Product with 1D Space Suppose $V$ and $W$ are inner product spaces over the complex numbers and $\operatorname{dim} V =1$. It seems to be true that
\begin{align}
(V\oplus W)^{\otimes N} \cong \bigoplus^N_{j=0} (V^{\otimes j}\otimes W^{\otimes (N-j)})\cong \bigoplus^N_{j=0} W^{\otimes j} \cong V\otimes \bigoplus^N_{j=0} W^{\otimes j}
\end{align}
where the isomorphism are in the sense of inner product spaces.
Two questions:

*

*Are the last two $\cong$ true?

*If 1) is true, then can we construct a unitary map from $(V\oplus W)^{\otimes N}$ to $V\otimes \bigoplus^N_{j=0} W^{\otimes j}$?

Edit: The above seems wrong as pointed out below. I meant to ask
\begin{align}
(V\oplus W)^{\otimes_s N} \cong \bigoplus^N_{j=0} (V^{\otimes_s j}\otimes_s W^{\otimes_s (N-j)})\cong \bigoplus^N_{j=0} W^{\otimes_s j} \cong V\otimes_s \bigoplus^N_{j=0} W^{\otimes_s j}
\end{align}
Then you have the binomial theorem. So, the first isomorphism is correct.
 A: Even the first isomorphism doesn't hold. For example,
$$ (V \oplus W)^{\otimes 2} = (V \oplus W) \otimes (V \oplus W) \cong V^{\otimes 2} \oplus \left( V \otimes W \right) \oplus \left( W \otimes V \right) \oplus W^{\otimes 2} $$
which is not isomorphic to
$$ W^{\otimes 2} \oplus \left( V \otimes W \right) \oplus V^{\otimes 2}. $$
To see this, let's say that the dimension of $W$ is $n$. Then the dimension of $(V \oplus W)^{\otimes 2}$ is $(n+1)^2$ while the dimension of $W^{\otimes 2} \oplus \left( V \otimes W \right) \oplus V^{\otimes 2}$ is $n^2 + n + 1 = (n+1)^2 - n$ as you are missing a factor of $W \otimes V$.

Regarding your question about the symmetric tensor product, let me use the more standard notation $S^j(V)$ for the $j$ symmetric power of a vector space $V$. Then we have the isomorphisms
$$ (1): \,\, V \cong S^j(V), \,\,\, v \mapsto v^j, \\
 (2): \bigoplus_{j=0}^n S^j(V) \otimes S^{N-j}(W) \cong S^N(V \otimes W), \\
\left(v_1 \dots v_j \right) \otimes \left( w_1 \dots w_{N-j} \right) \mapsto v_1 \dots v_j w_1 \dots w_{N-j}$$
and hence we can build an isomorphism
$$ S^N(V \oplus W) \cong \bigoplus_{j=0}^N S^j(V) \otimes S^{N-j}(W) \cong \bigoplus_{j=0}^N V \otimes S^{N-j}(W) \cong V \otimes \bigoplus_{j=0}^N S^j(W). $$
Explicitly, the isomorphism $V \otimes \bigoplus_{j=0}^N S^j(W) \mapsto S^n(V \otimes W)$ is given (on decomposable elements) as
$$ v \otimes \left( \sum_{j=0}^N w^j_1 \dots w^j_j \right) \mapsto \sum_{j=0}^N v^j w_1^j \dots w_j^j.$$
As to whether this isomorphism is an isometry, it really depends on how you define the inner product on the symmetric power. One natural way to do it is to define
$$ \left< v_1 \dots v_k, v_1' \dots v_k' \right> = \sum_{\sigma \in S_k} \prod_{i=1}^k \left< v_i, v_{\sigma(i)}' \right>. $$
With respect to this definition, the isomorphism $(2)$ becomes an isometry (when using the functorial inner products on the direct sum and tensor product) but $(1)$ is not and has to be replaced with $v \mapsto \frac{1}{\sqrt{j!}} v$ so the resulting isometry is
$$ v \otimes \left( \sum_{j=0}^N w^j_1 \dots w^j_j \right) \mapsto \sum_{j=0}^N \frac{v^j w_1^j \dots w_j^j}{j!}. $$
Your definition of the inner product might vary by a factor and then you need to compensate for that factor.
