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Homework Help: Complexification of subspace

  1. Aug 28, 2010 #1

    julian

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    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
     
  2. jcsd
  3. Aug 29, 2010 #2

    lanedance

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    had a quick look & think it should work both ways

    [tex]\Rightarrow[/tex]
    [tex] p \in F^{\perp}[/tex]
    [tex] p_c \in (F^{\perp})_c, \ \ p_c = p_1 + ip_2, \ \ p_1, p_2 \in F^{\perp} [/tex]

    now take
    [tex] f_c \in F_c, \ \ f_c = f_1 + i f_2, \ \ f_1, f_2 \in F [/tex]

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

    so
    [tex] p_c \in (F_c)^{\perp}[/tex]
     
  4. Aug 29, 2010 #3

    lanedance

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    and the other
    [tex] \Leftarrow[/tex]

    [tex] f_c \in F_c, \ \ f_c = f_1 + i f_2, \ \ f_1, f_2 \in F [/tex]

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

    [tex]q \in V_c, \ \ q = q_1 + iq_2, \ \ q_1,q_2 \in V [/tex]

    [tex] w[q,f_c]=0 \ \implies \ w[q_1, f_1] = 0 \ \implies \ q_1 \in F^{\perp} [/tex]
     
    Last edited: Aug 29, 2010
  5. Aug 29, 2010 #4

    julian

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    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 ?
     
  6. Aug 29, 2010 #5

    lanedance

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    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)
     
  7. Aug 30, 2010 #6

    julian

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    I'm not sure you can put f_2 = 0 - it's not necessarily an element of F
     
  8. Aug 30, 2010 #7

    lanedance

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    F is a subspace, so must contain the zero vector by def'n
     
  9. Aug 31, 2010 #8

    julian

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    Are you assuming the subspace is a Vector subspace?

    thanks
     
  10. Aug 31, 2010 #9

    lanedance

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    yeah thats generally what's meant by a subspace of a vector space...
     
    Last edited: Aug 31, 2010
  11. Sep 15, 2010 #10

    julian

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    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 dont 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.
     
  12. Sep 15, 2010 #11

    julian

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    sorry I meant


    (G_R)_C = G
     
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