Let $G$ be a reductive algebraic group over $k$, and $V\_i$ be (finite-dimensional) representations of $G$. Are the irreducible components of $\bigotimes\_i V\_i$ always generated by indecomposable vectors? That is, vectors of the form $v\_1 \otimes \cdots \otimes v\_n$ where $v\_i \in V\_i$. 



Are irreducible subrepresentations of a tensor product always generated by indecomposable vectors?
It seems unlikely, given that already the assertion is not literally true for the tensor product of two copies of the standard representation $V$ of $G=SL\_2(\mathbb R)$. Namely, with $v$ a highest-weight vector in $V$, and $w$ a lowest-weight vector in $V$, with $Lv=w$ with lowering operator $L$, in $V\otimes V$, by Leibniz' rule,
$$
L(v\otimes w) \;=\; Lv\otimes w + v\otimes Lw
\;=\; w\otimes w + 0 \;=\; w\otimes w
$$
Thus, looking at the two weight-$0$ vectors in $V\otimes V$, neither $v\otimes w$ nor $w\otimes v$ is annihilated by the lowering operator. We know that $V\otimes V\approx {\mathrm{Sym}^2}\oplus {\mathrm{trivial}}$, and that the weight of the trivial representation is $0$. Indeed, $L(v\otimes w-w\otimes v)=0$, so $v\otimes w-w\otimes v$ is the vector generating the trivial repn inside the tensor product.


If $v\otimes w - w\otimes v$ were equal to $(av+nw)\otimes (cv+dw)$, upon multiplying out $ac=0$, so the coefficients of both $v\otimes w$ and $w\otimes v$ are $0$, and this is impossible.


