How to show that $\hbox{Gr}(2,\mathbb{C}^5)$ is homeomorphic to $\hbox{Gr}(3,\mathbb{C}^5)$ by showing that they are given by the same Plücker relations?

I'm trying to understand Grassmannian and Plücker relationship and I'm having trouble grasping the basic idea.


1 Answer 1


Question: "I'm trying to understand Grassmannian and Plücker relationship and I'm having trouble grasping the basic idea."

Answer: If $k$ is a field and $W \subseteq V$ is an $m$-dimensional $k$-vector subspace of an $n$-dimensional $k$-vector space it follows grassmannian $\mathbb{G}(m,V)$ is a parameter space, parametrizing $m$-dimensional sub vector spaces of $V$. The grassmannian $\mathbb{G}(m,V)$ is a smooth projective algebraic variety over $k$. If $k$ is the field of real numbers (or complex numbers) it follows $\mathbb{G}(m,V)$ is a real smooth manifold (or a complex projective manifold).

Example: If $m:=1, n:=d+1$ it follows $\mathbb{G}(m,V) \cong \mathbb{P}^d_k$ is projective $d$-space over $k$, parametrizing lines in $V$.

There is always an exact sequence

$$0 \rightarrow W \rightarrow V \rightarrow V/W \rightarrow 0$$

of $k$-vector spaces and dualizing this sequence you get an exact sequence

$$0 \rightarrow (V/W)^* \rightarrow V^* \rightarrow W^* \rightarrow 0$$

and $\dim_k((V/W)^*)=n-m.$ Hence vieweing $[W] \in \mathbb{G}(m,V)$ as a "point" you get canonically a "point" $[(V/W)^*] \in \mathbb{G}(n-m,V^*)$ and this gives an isomorphism of varieties

$$\mathbb{G}(m,V) \cong \mathbb{G}(n-m, V^*).$$

Since $\dim_k(V)=\dim_k(V^*)$ and if you ignore that there is no "canonical isomorphism" $V \cong V^*$ as $k$-vector spaces, the result "follows" for any $m,n$.

In your case if $V:=\mathbb{C}^5$ and you are able to give a "canonical isomorphism" $V \cong V^*$ you get a "canonical isomorphism"

$$ \mathbb{G}(2, V) \cong \mathbb{G}(3,V^*) \cong \mathbb{G}(3, V).$$

If $V:=k\{e_1,..,e_5\}$ and $V^*:=k\{x_1,..,x_5\}$ with $x_i:=e_i^*$, it follows at the level of ideals

$$I(\mathbb{G}(2,V)) \subseteq Sym_k^*((\wedge^2 V)^*):=k[y_1,..,y_{10}]$$


$$I(\mathbb{G}(3,V^*)) \subseteq Sym_k^*((\wedge^3 V^*)^*):=k[(u_1)^*,..,(u_{10})^*]$$

where $\wedge^3 V^* \cong k\{u_1,..,u_{10} \}$. Notice that the Plucker embedding gives embeddings

$$\mathbb{G}(2,V) \subseteq \mathbb{P}((\wedge^2 V)^*)$$


$$\mathbb{G}(3,V^*) \subseteq \mathbb{P}((\wedge^3 V^*)^*)$$

but $\dim_k(\wedge^2 V) =\dim_k(\wedge^3 V^*)$ hence $\mathbb{P}((\wedge^2 V)^*) \cong \mathbb{P}((\wedge^3 V^*)^*)$. Hence strictly speaking: the two ideals live in different rings: The coordinate rings of $\mathbb{P}((\wedge^2 V)^*)$ and $\mathbb{P}((\wedge^3V^*)^*)$ differ and it is not clear how to relate different choices of coordinates. You may choose an isomorphism and calculate the two ideals and compare. Projective space $\mathbb{P}^9$ has many automorphisms and choosing coordinates gives you two embeddings

$$\mathbb{G}(2,V), \mathbb{G}(3,V^*) \subseteq \mathbb{P}^9$$

and you could end up having chosen coordinates in such a way that there is an automorphism $g \in Aut_k(\mathbb{P}^9)$ with $ \mathbb{G}(3,V^*)=g(\mathbb{G}(2,V))$.

In the previously mentioned book you find an algorithm giving generators for the above ideals and then you can compare. $Aut_k(\mathbb{P}^n_k) \cong PGL(n,k)$ and if you make an arbitrary choice of coordinates it is very likely you end you end up with a non-trivial automorphism $g \in PGL(n,k)$. You must make a "smart" choice.

Algebraically independent Plücker relations

Note: In Fulton/Harris "Representation theory: A first course" you find a description of the relation between $\wedge^m(V^*)$ and $(\wedge^m V)^*$ and "dual bases". When you dualize exterior products and symmetric products there are explicit non-trivial formulas for the "dual basis" - this choice is "natural". This may help. There is a canonical map

$$\wedge^2 V \otimes \wedge^3 V \rightarrow \wedge^5 V \cong \mathbb{C}$$

inducing a canonical isomorphism $\wedge^2 V \cong (\wedge^3 V)^*$. There is a "natural map"

$$ \phi: \wedge^n(V^*) \cong (\wedge^n V)^* $$

defined by

$$\phi(\phi_1 \wedge \cdots \wedge \phi_n)(v_1 \wedge \cdots \wedge v_n):=$$


This gives a "natural isomorphism" $\wedge^2 V \cong \wedge^3 (V^*)$ and a "natural isomorphism" $\mathbb{P}((\wedge^2 V)^*) \cong \mathbb{P}((\wedge^3 (V^*)^*)$. It could be this is a "smart choice" - this must be checked.

Note 1: This construction generalize: There is a "natural" isomorphism $\phi:\mathbb{P}((\wedge^k V)^*) \cong \mathbb{P}((\wedge^{n-k}V^*)^*)$ and you may ask if the map $\phi$ induce an isomorphism

$$\phi:\mathbb{G}(k,V) \cong \mathbb{G}(n-k, V^*).$$

Note 2: You may also consider the parabolic subgroup $P \subseteq SL(V)$ fixing the subspace $W$ and there is an isomorphism $\mathbb{G}(2,V) \cong SL(V)/P$. You may do something similar with $\mathbb{G}(3,V^*)$ and try to write down an explicit isomorphism of schemes

$$SL(V)/P \cong SL(V^*)/P'$$

where $P' \subseteq SL(V^*)$ is the subgroup fixing $(V/W)^*$. There is an isomorphism $SL(V) \cong SL(V^*)$ and an abstract isomorphism $P \cong P'$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .