I have a general question, and then the specific version of that question I need for research. All vector spaces over $\mathbb{C}$.


Grassmanians of planes
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The $(2,n)$-Grassmannian, denoted $Gr(2,n)$, is the space of all 2 dimensional subspaces of $n$ dimensional space. This is naturally a complete complex variety. Its variety structure is usually realized through the **Plucker embedding**


$$Gr(2,n)\hookrightarrow \mathbb{P}^{\binom{n}{2}}$$


This starts with a space $V\subset \mathbb{C}^n$, chooses basis vectors $v\_1,v\_2$, and then maps $V$ to the point in $\mathbb{P}^{\binom{n}{2}}$ with homogenous coordinate given by all $2\times 2$-minors of the $2\times n$ matrix with rows $v\_1,v\_2$.


For $i,j\in [n]$, the **Plucker coordinates** $x\_{ij}$ is the $\{i,j\}$th homogeneous coordinate on $\mathbb{P}^{\binom{n}{2}}$; therefore, the composition with the Plucker embedding corresponds to the $ij$th minor. The image of this map is defined by the homogeneous **Plucker relations**. For all, $i<j<k<l$, one has
$$ x\_{ik}x\_{jl}=x\_{ij}x\_{kl}+x\_{il}x\_{jk}$$
Let $\mathcal{O}Gr(2,n)$ denote the *homogeneous coordinate ring*, this is the graded algebra generated by the Plucker coordinates, modulo the Plucker relations.


The group $PGL\_n(\mathbb{C})$ acts transitively on $Gr(2,n)$, by its action on $\mathbb{C}^n$. Therefore, $Gr(2,n)$ is a homogeneous space. From this, it follows immediately that $Gr(2,n)$ is a smooth, irreducible variety.


Smoothness of hypersurfaces
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Any homogeneous element $f$ in $\mathcal{O}Gr(2,n)$ defines a hypersurface $V\_f$ in $Gr(2,n)$. My general question is:

>  How can one effectively check if $V\_f$ is smooth? 




As an example of the problem, the main general technique I know for showing a variety is smooth is to embed it in affine or projective space with codimension $c$, and then check the $c\times c$ minors of the Jacobian matrix of the defining ideal. The problem with this approach here is that the codimension of the Plucker embedding is quite large in general, and the minors rapidly become unwieldy.


My specific concern
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Outside curiosity, I am not interested in the problem for general hypersurfaces. The specific hypersurface I have in mind is defined by $f=x\_{ij}-x\_{kl}$, the difference of two Plucker coordinates. This is a frustratingly simple variety, but I am having trouble saying things about it. Is it smooth? Is it irreducible?


More generally, I can look at the subscheme defined by intersecting several hypersurfaces of this form. Is it smooth? Is it irreducible?



Smoothness of hypersurfaces in Grassmannians
Here is the answer to your specific question (and, in fact, something a little more general):


The hypersurface defined by $f=x\_{ij}-x\_{kl}$ (assuming that $i$, $j$, $k$, and $l$ are distinct) is smooth if $n\le 5$ and singular for $n>5$. The singular locus consists of the $2$-planes that lie in the codimension $4$ subspace that consists of $0$ plus the vectors $v$ such that pairing $v$ with *any* linearly independent vector generates a $2$-plane that still lies in the hypersurface.


The way to think about this more geometrically is this: Take any nonzero $2$-form $\Omega$ on $\mathbb{C}^n$. Consider the set $Z\_\Omega\subset Gr(2,n)$ that consists of the $2$-planes on which $\Omega$ vanishes. This is a hypersurface in $Gr(2,n)$ of exactly the kind you are considering in your specific problem. Let $N\subset\mathbb{C}^n$ be the *null space* of $\Omega$, i.e., the subspace consisting of the vectors $v\in\mathbb{C}^n$ such that $\Omega(v,w) = 0$ for all $w\in\mathbb{C}^n$. Then a $2$-plane $V$ is a singular point of $Z\_\Omega$ if and only if $V$ is a subspace of $N$. 


In your particular case, $\Omega = dz^i\wedge dz^j - dz^k\wedge dz^l$, so $N$ is the subspace defined by $z^i = z^j = z^k = z^l = 0$. 


In the more general case, the codimension of $N$ is twice the largest integer $\rho$ such that $\Omega^\rho\not=0$.


These claims are easily verified in a chart on $Gr(2,n)$.


Addendum: I forgot to answer the question about whether the hypersurface $Z\_\Omega$ is irreducible. It is. The reason is that the smooth locus is connected, as follows without too much difficulty from the above description.


Also, you asked whether the intersection of a number of these 'linear' hypersurfaces is always connected. The answer to this is 'no'. For a simple example, take the locus defined by $x\_{13}=x\_{14}=x\_{23}=x\_{24}=0$ in $Gr(2,4)$. This consists of two points, a pair of 2-planes in general position in $\mathbb{C}^4$.


In general, deciding whether such an intersection is smooth or connected (or even nonempty) is not easy. In the theory of exterior differential systems, there is a criterion that is sufficient for a point in $Gr(2,n)$ to be smooth that is often useful, even though it is far from necessary. I'll describe it briefly here, but you can get more information in our book *Exterior Differential Systems*. 


For notational simplicity, let me denote $\mathbb{C}^n$ by $W$. Let $\Sigma\subset \Lambda^2(W^\ast)$ be a linear subspace, and let $Z\_\Sigma\subset Gr(2,W)$ denote the set of $2$-planes $V\subset W$ such that all of the elements of $\Sigma$ vanish on $V$. I want to define the locus $Z^o\_\Sigma\subset Z\_\Sigma$ of *ordinary* elements, and these will be smooth points of $Z\_\Sigma$. To do this, for each $w\in W$, consider the vector space $H(w)\subset W$ consisting of all the vectors $v\in W$ such that $\Omega(v,w)=0$ for all $\Omega\in\Sigma$. Say that $w$ is *$\Sigma$-regular* if the dimension of $H(w)$ is minimal among all $w\in W$. (The $\Sigma$-regular elements in $W$ form a Zariski-open subset of $W$.) Say that $V\in Z\_\Sigma$ is *$\Sigma$-ordinary* if $V$ contains a $\Sigma$-regular vector. 


It is not hard to show that the set $Z^o\_\Sigma\subset Z\_\Sigma$ consisting of $\Sigma$-ordinary elements consists of smooth points of $Z\_\Sigma$, and its closure is an irreducible component of $Z\_\Sigma$. However, $Z^o\_\Sigma$ might well be empty, even though $Z\_\Sigma$ is nonempty (and could even be smooth). For the purposes of exterior differential systems, though, $Z^o\_\Sigma$ turns out to be the most interesting part of $Z\_\Sigma$ (when it is nonempty).


