State orthogonal to a symmetric state must be antisymmetric I am working out the exchange symmetry of the eigenstates of the total angular momentum operator of a system of two spin-1 bosons.
I know that there must be a quintet, triplet, and a singlet state.
The highest state of the quintet is $|\uparrow\rangle |\uparrow\rangle$ and is symmetric under particle exchange.
Applying the lowering operator in this state generates the entire quintet. Since the lowering operator is symmetric itself, all states in the quintet are symmetric.
Using the orthogonality of states, one can deduce the highest state of the triplet. Successively applying the lowering operator generates the entire triplet.
My problem is: How does one show that the triplet states are antisymmetric, given that the quintet states are symmetric.
Phrased more generally perhaps, if the state $|\Psi\rangle $ is symmetric under particle exchange and state $|\Phi\rangle$ is such that $\langle \Psi | \Phi \rangle=0$, can one show that $|\Phi\rangle$ is antisymmetric?
 A: In fact as stated
$$\langle \Phi\vert \Psi\rangle =0 \tag{1}
$$ does not imply that one is symmetric and the other antisymmetric.  The coupling $1\otimes 1=2\oplus 1\oplus 0$ and the state with $L=0$ is orthogonal to all the $L=2$ states yet it is also symmetric, v.g. with $\vert \Phi\rangle=\vert 2,0\rangle$ and $\vert\Psi\rangle=\vert 0,0\rangle$ then both are symmmetric but (1) still,holds.
A: You are overthinking it; you didn't do this in class?
For your specific example, you see it. Ignoring normalizations,
$ |\uparrow\rangle |\uparrow\rangle$ is lowered by $J_-$
to  $|\uparrow\rangle | 0\rangle+  |0\rangle |\uparrow\rangle$, and so on, as you indicated. But this state is orthogonal to  $|\uparrow\rangle | 0\rangle -  |0\rangle |\uparrow\rangle$, which is antisymmetric, so it is annihilated by $J_+$, so it's the highest spin state of the triplet.
Lowering it once and twice will likewise produce antisymmetric states.
The singlet, $|\uparrow\rangle | \downarrow\rangle +  | \downarrow\rangle|\uparrow\rangle  -  |0\rangle | 0\rangle$, will be symmetric.
