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A basic way to describe the $3$-dimensional standard irreducible representation of $S_4$ is the following (I'll do it over $\mathbb{Q}$): Let $z_1, z_2, z_3, z_4$ be the standard basis of $\mathbb{Q}^4$. The symmetric group $S_4$ acts on $\mathbb{Q}^4$ by permuting the basis vectors. It acts trivially on the space $U$ generated by the element $z_1 + \cdots + z_4$. Hence we may describe the standard $3$-dimensional irrep of $S_4$ as the space $$\bar{U} := \mathbb{Q}^4 / U \cong \{ v = (v_1, v_2, v_3, v_4) \in \mathbb{Q}^4 \mid v_1 + \cdots + v_4 = 0 \}.$$

Now the other $3$-dimensional irrep of $S_4$ is the product of the standard irrep and the sign irrep.

Is there a simple way to describe this signed standard irrep of $S_4$ similar to the one above for the standard irrep?

Starting as above with $\mathbb{Q}^4$, we'd have to look at the action of $S_4$ on $\mathbb{Q}^4$ by permutation of the basis elements $z_1, \ldots, z_4$ times the determinant of the respective element of $S_4$, i.e. for $\sigma \in S_4$: $$\sigma(k_1 z_1 + \cdots + k_4 z_4) = \det(\sigma) \left( k_1 z_{\sigma(1)} + \cdots + k_4 z_{\sigma(4)} \right).$$ But I think there is no subspace of $\mathbb{Q}^4$ on which this action of $S_4$ is trivial, right? (compare $\bar{U}$ for the standard irrep)

PS: I realize that the usual way to look at this is to take the tensor product $\bar{U} \otimes W$ where $W$ is the $1$-dimensional sign representation of $S_4$. What I really want though (I guess) is to see how the signed standard irrep of $S_4$ can be embedded in $\mathbb{Q}^4$ (or viewed as a quotient of $\mathbb{Q}^4$) in a simple way similar to the standard irrep.

Thanks in advance for any help!

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    No, $\overline {U}\otimes W $ cannot be embedded into $\mathbb {Q}^4$. – 2017-01-13
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    @darijgrinberg Could you provide a reference or give an argument for this? – 2017-01-13
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    Most texts on the representation theory of symmetric groups should land you in a place from where you can reach it on foot; I don't know one which spells it out. Basically, representations of $S_4$ are semisimple, meaning that they can decomposed into a direct sum of irreducibles, and the isomorphism classes of these irreducibles are uniquely determined; moreover, each irreducible representation appearing as a submodule or a quotient or even a subquotient of some representation $P$ must be isomorphic to a direct summand of $P$. Thus, we only need to know that $\overline{U} \otimes W$ is ... – 2017-01-13
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    ... not a direct summand of $\mathbb{Q}^4$. The irreducible direct summands of $\mathbb{Q}^4$, however, are easy to find: They are $U$ and its orthogonal complement isomorphic to $\overline{U}$. So there is "no place" for other irreducible representations, such as $\overline{U} \otimes W$. The same phenomenon holds in higher degrees: The irreducible components of the standard representation $\mathbb{Q}^n$ of the symmetric group $S_n$ are the trivial representation (consisting of all vectors $\left(v,v,\ldots,v\right)$) and ... – 2017-01-13
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    ... its orthogonal complement (consisting of all vectors $\left(v_1,v_2,\ldots,v_n\right)$ with $v_1+v_2+\cdots+v_n=0$). For $n \geq 4$, none of these two is isomorphic to its tensor product with the sign representation. – 2017-01-13
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    I'm confused by @darijgrinberg 's comments. The action you have defined on $\mathbb{Q}^4$ near the end of your post contains a copy of the sign representation. The complement (or quotient) yields the 3-dimensional representation you are looking for. – 2017-01-13
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    Ah, you want to embed it in *that* $\mathbb{Q}^4$ ? I think it's worthwhile to clean up the original post somewhat, giving non-ambiguous names to representations. – 2017-01-13

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