Complexification of subspace

[julian]

Let F be a subspace of a real vector space V and let

G \subset V_C

i.e. a subspace of its complexification. Define the real subspace of G by


G_R := G \cap V.


There is a symplectic form w[u,v]. The annihilator subspace F^perp of V is defined by


F^perp = {v \in V : w[u,v] = 0 for all u \in F}


The annihilator subspace G^perp of V_C is defined by


G^perp = {v \in V_C : w[u,v] = 0 for all u \in G}.


Then the following results hold:


(F^perp)_C = (F_C)^perp,


It is easy to see that (F^perp)_C subset of (F_C)^perp. Could somone prove it the other way? I'm thinking F_C is larger than F so the annihilator condition is more restrictive?

thanks

 

[lanedance]

had a quick look & think it should work both ways

$$\Rightarrow$$
$$p \in F^{\perp}$$
$$p_c \in (F^{\perp})_c, \ \ p_c = p_1 + ip_2, \ \ p_1, p_2 \in F^{\perp}$$

now take
$$f_c \in F_c, \ \ f_c = f_1 + i f_2, \ \ f_1, f_2 \in F$$

$$w[p_c,f_c] = w[ f_1 + if_2, p_1 + ip_2 ] = 0$$

so
$$p_c \in (F_c)^{\perp}$$

 

[lanedance]

and the other
$$\Leftarrow$$

$$f_c \in F_c, \ \ f_c = f_1 + i f_2, \ \ f_1, f_2 \in F$$

$$q \in (F_c)^{\perp}, \ \ w[f_c,q] = 0 , \ \ \forall f_c \in F_c$$

$$q \in V_c, \ \ q = q_1 + iq_2, \ \ q_1,q_2 \in V$$

$$w[q,f_c]=0 \ \implies \ w[q_1, f_1] = 0 \ \implies \ q_1 \in F^{\perp}$$

 

[julian]

Thank you.


to prove the very last part do you put f_1=f_2 which implies from real and imaginary parts of w [q,f_c] are


w [q_1 + q_2 , f_1] =0

w[q_1 - q_2, f_1] = 0

adding gives

w[q_1,f_1] = 0 ?

 

[lanedance]

no i'ev assumed w is a linear operator
w[q_1 + i q_2, f_1] = w[q_1, f_1] + w[i q_2, f_1] = 0

so it follows that each part must be zero (though you may need to be careful with conjugates)

 

[julian]

I'm not sure you can put f_2 = 0 - it's not necessarily an element of F

 

[lanedance]

F is a subspace, so must contain the zero vector by def'n

 

[julian]

Are you assuming the subspace is a Vector subspace?

thanks

 

[lanedance]

yeah that's generally what's meant by a subspace of a vector space...

 

[julian]

A subspace will be a vector space if it is closed under addition and multiplication by the field, but what if these conditions are not imposed?


I have another question (I don't know if it has a relation to the above). I mentioned a G at the begining. In the book he also says that if G=G^* then (G_R)_C. I know that we cannot have $z w = w^* in general and maybe that has something to do with it.

 

[julian]

sorry I meant


(G_R)_C = G