- #1
Kreizhn
- 743
- 1
Hey all,
I have a quick question that should hopefully be simple to answer.
Consider a the space of [itex] n \times n [/itex] matrices over [itex] \mathbb C[/itex] given by [itex] M_n(\mathbb C) [/itex]. In order to properly consider this as a real matrix, we have to embed [itex] M_n(\mathbb C) \to M_{2n}(\mathbb R) [/itex], and I can give some books that cite this. In order to do this, we use one of the following methods:
Method 1
Define the map [itex] \rho: \mathbb C \to M_2(\mathbb R) [/itex] by
[tex] \rho(x+iy) = \begin{pmatrix} x & -y \\ y & x \end{pmatrix} [/tex]
which actually gives an identification of the complex numbers with a subspace of [itex] M_2(\mathbb R) [/itex]. Let [itex] Z \in M_n(\mathbb C) [/itex] be given by the components [itex] Z=[Z]_{ij}[/itex]. Then define [itex] \rho_n: M_n(\mathbb C) \to M_{2n}(\mathbb R) [/itex] as [itex] \rho_n(Z) = [\rho(Z_{ij})]_{ij} [/itex]. That is, we have just made each complex entry of Z into a 2x2 matrix.
Method 2
Similar to above, let [itex] Z = X+iY \in M_N(\mathbb C) [/itex] where [itex] X,Y \in M_n(\mathbb R) [/itex] and define [itex] \rho': M_n(\mathbb C) \to M_{2n}(\mathbb R) [/itex] by
[tex] \rho'(Z) = \begin{pmatrix} X & -Y \\ Y & X \end{pmatrix} [/tex]
where this is done in block-matrix form.
Now both of these methods are related, and the reason for the relation comes from the two different definitions of symplectic structures. However, this is not what I'm interested in. We notice that [itex] M_{2n}(\mathbb R) [/itex] is a [itex] 4n^2 [/itex] dimensional space and in particular we can identify it with [itex]\mathbb R^{4n^2} [/itex]. What if instead of using one of these embeddings, we instead wrote [itex] Z = X + i Y [/itex] and visualized it as an element of [itex] M_{n,2n}(\mathbb R) [/itex] with
[tex] Z = \begin{pmatrix} X \\ Y \end{pmatrix} [/tex]
which has dimension [itex] 2n^2 [/itex] and so we identify it with [itex] \mathbb R^{2n^2} [/itex]. Why do we prefer using one of the previous two methods if they give a higher dimensional space? Is it for some reason like "multiplication is properly preserved" or something along those lines?
Edit: Something in my head says that from a topological point of view, there's probably not a big difference. However, it would be something like "We want to preserve the ring structure of [itex] \mathbb C [/itex]."
I have a quick question that should hopefully be simple to answer.
Consider a the space of [itex] n \times n [/itex] matrices over [itex] \mathbb C[/itex] given by [itex] M_n(\mathbb C) [/itex]. In order to properly consider this as a real matrix, we have to embed [itex] M_n(\mathbb C) \to M_{2n}(\mathbb R) [/itex], and I can give some books that cite this. In order to do this, we use one of the following methods:
Method 1
Define the map [itex] \rho: \mathbb C \to M_2(\mathbb R) [/itex] by
[tex] \rho(x+iy) = \begin{pmatrix} x & -y \\ y & x \end{pmatrix} [/tex]
which actually gives an identification of the complex numbers with a subspace of [itex] M_2(\mathbb R) [/itex]. Let [itex] Z \in M_n(\mathbb C) [/itex] be given by the components [itex] Z=[Z]_{ij}[/itex]. Then define [itex] \rho_n: M_n(\mathbb C) \to M_{2n}(\mathbb R) [/itex] as [itex] \rho_n(Z) = [\rho(Z_{ij})]_{ij} [/itex]. That is, we have just made each complex entry of Z into a 2x2 matrix.
Method 2
Similar to above, let [itex] Z = X+iY \in M_N(\mathbb C) [/itex] where [itex] X,Y \in M_n(\mathbb R) [/itex] and define [itex] \rho': M_n(\mathbb C) \to M_{2n}(\mathbb R) [/itex] by
[tex] \rho'(Z) = \begin{pmatrix} X & -Y \\ Y & X \end{pmatrix} [/tex]
where this is done in block-matrix form.
Now both of these methods are related, and the reason for the relation comes from the two different definitions of symplectic structures. However, this is not what I'm interested in. We notice that [itex] M_{2n}(\mathbb R) [/itex] is a [itex] 4n^2 [/itex] dimensional space and in particular we can identify it with [itex]\mathbb R^{4n^2} [/itex]. What if instead of using one of these embeddings, we instead wrote [itex] Z = X + i Y [/itex] and visualized it as an element of [itex] M_{n,2n}(\mathbb R) [/itex] with
[tex] Z = \begin{pmatrix} X \\ Y \end{pmatrix} [/tex]
which has dimension [itex] 2n^2 [/itex] and so we identify it with [itex] \mathbb R^{2n^2} [/itex]. Why do we prefer using one of the previous two methods if they give a higher dimensional space? Is it for some reason like "multiplication is properly preserved" or something along those lines?
Edit: Something in my head says that from a topological point of view, there's probably not a big difference. However, it would be something like "We want to preserve the ring structure of [itex] \mathbb C [/itex]."