Schur functors for $\mathfrak{S}_3$

I have been trying to calculate the explicit images of the Schur functors for the action of $$\mathfrak{S}_3$$ on $$V^{\otimes 3}$$ where $$V$$ is some vector space, for the sake of concreteness of dimension 2.

$$\mathfrak{S}_3$$ has 3 irreducible representations, namely the trivial, alternating and standard. Using their characters and the character of the representation on $$V^{\otimes 3}$$, it is easy to see that their multiplicities in the decomposition of $$V^{\otimes 3}$$ into irreducibles are 4, 0 and 2 respectively.

Now, the Schur functor associated to the partition $$(3,0,0)$$ is precisely the symmetrisation functor, and its image is $$\mathrm{Sym}^3 V$$, which has dimension 4, as is to be expected. On the other hand, the image of the projector associated to the partition $$(1,1,1)$$ is the antisymmetrisation functor, and its image is 0, which makes sense since $$V$$ has no third exterior power.

What puzzles me is the functor corresponding to the standard representation. The Young symmetriser in this case is $$1 + (12) - (13) - (123)$$. While doing the calculation I was expecting its image to be 4 dimensional, which would then further decompose into the copies of the standard representation predicted by the characters. But, unless my calculation is wrong, the subspace that I get is only of dimension 2. If $$a$$ and $$b$$ are a basis for $$V$$, then the images of $$a\otimes a \otimes a$$ and $$b\otimes b \otimes b$$ both vanish. All three products of two $$a$$s and one $$b$$ get sent to multiples of $$b\otimes a \otimes a - a \otimes a \otimes b$$, and exchanging $$a$$ and $$b$$, the image of the products of two $$b$$s and one $$a$$ are multiples of $$a\otimes b \otimes b - b \otimes b \otimes a$$. This space is not even a subrepresentation.

I would have hoped to get instead the span of $$b\otimes a \otimes a - a \otimes a \otimes b$$ and $$a\otimes a \otimes b - a \otimes b \otimes a$$, which is isomorphic to the standard representation, and then the corresponding subspace under the exchange of $$a$$ and $$b$$. Is this correct? Is the image of a Young symmetriser not a representation or am I missing something?

Your $$a \otimes a \otimes a$$ vanishes under this Young Symmetrizer (which corresponds to the space $$\mathbb{S}_{2,1} V$$, see https://en.wikipedia.org/wiki/Schur_functor) since it belongs to $$Sym^3V$$. It won't vanish under the symmetrizer for the partition $$(3, 0, 0)$$, which is just the sum $$\sum_{\sigma \in S_3} \sigma$$.
Finally, the space you obtained ($$Span \{ b \otimes a \otimes a - a \otimes a \otimes b, \ a \otimes b \otimes b - b \otimes b \otimes a \}$$) Is a representation of $$GL(2, \mathbb{C})$$ (you can check that this subspace is invariant under the operation $$U\otimes U\otimes U$$). On the other hand, $$Span \{ b \otimes a \otimes a - a \otimes a \otimes b, \ a \otimes a \otimes b - a \otimes b \otimes a \}$$ is a representation of $$S_3$$.