# Grassmannian $\hbox{Gr}(2,\mathbb{C}^5)$ is homeomorphic to $\hbox{Gr}(3,\mathbb{C}^5)$

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.

• Sep 12, 2021 at 12:59

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

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

and

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

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

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

and

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

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

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

$$\det(\phi_j(v_i)).$$

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 $$P \cong P'$$.