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Is the dimension of the projective space $\mathbb{P}^{{n \choose k} -1}$ into which we embed the Grassmannian $G(k,n)$ of $k$-planes in $n$-space minimal? In other words, is the Grassmannian variety (ever) contained in a Plücker hyperplane section?

Sorry, I wasn't able to find a reference for this.

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The Grassmannian in its Plücker embedding spans the space. The space $\mathbb{P}^{\binom{n}{k}-1}$ of alternating tensors is spanned by the simple wedges (also called decomposable) $v_1 \wedge \dotsb \wedge v_k$, i.e., elements of the Grassmannian. So, no, in this embedding the Grassmannian is not contained in any hyperplane.

But that does not mean it's the minimum possible ambient dimension in which the Grassmannian can be embedded. You can project from a point outside of the secant variety of the Grassmannian to get an embedding in a smaller dimension.

I don't know what is the smallest dimension a Grassmannian can be embedded into.

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    $\begingroup$ In slightly fancier algebraic geometry terms: maps of $X$ to projective space $\mathbb{P}(V)$ come from a map from $V^*$ into the sections of a line bundle on $X$. The image is contained in a hyperplane if this map has a non-trivial kernel (with the hyperplane determined by an element of the kernel). The Plücker embedding comes from taking all the sections of the determinant bundle of the tautological bundle whose fiber at a subspace $U$ is the dual $U^*$, and thus the map is obviously injective. $\endgroup$
    – Ben Webster ♦
    Commented May 31, 2016 at 5:07

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