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Embedding {$U(2n)\supset U(n)\times U(n)$}

Consider the compact Lie groups: orthogonal group {$O(n)$}, unitary group {$U(n)$}, compact symplectic group {$Sp(n)$}.

• Orthogonal matrix preserves the inner product of vectors of real numbers.
• Unitary matrix preserves the inner product of vectors of complex numbers.
• Compact symplectic matrix preserves the standard Hermitian form on vectors of quaternions.

I am investigating the chains of Lie group embeddings:

{$$U(2r)\supset U(r)\times U(r) \supset U(r)$$}

This manifests 2-fold complex Bott periodicity.

{$$O(16r)\supset U(8r)\supset Sp(4r)\supset Sp(2r)\times Sp(2r)\supset Sp(2r)\supset U(2r) \supset O(2r)\supset O(r)\times O(r) \supset O(r)$$}

This manifests 8-fold real Bott periodicity.

How do we generate these chains? Consider a linear complex structure {$J$}. In some basis, it can be written

{$$J=\begin{pmatrix}0 & -1 & & & \\ 1 & 0 & & 0 & \\ & & \ddots & & \\ & 0 & & 0 & -1 \\ & & & 1 & 0\end{pmatrix}$$}

Then the 2-fold Lie group embedding is generated by retaining what commutes with {$iJ_1$} and then a mutually anticommuting {$iJ_2$}.

That is, {$(iJ_1)(iJ_2)=-(iJ_2)(iJ_1)$} or simply {$J_1J_2=-J_2J_1$}.

The 8-fold Lie group embedding is generated by retaining what commutes with mutually anticommuting {$J_1,J_2,\dots J_k$}.

When we have an operator {$K^2=I$} on {$V$}, our vector space splits {$V=V_+\oplus V_-$}

Note that {$J_1^2=–I$}. Whereas {$(iJ_1)^2=i^2J_1^2=(-1)(-1)=+1$}. The operator {$K=iJ_1$} satisfies the equation {$K^2-I=0$}.

If {$\lambda$} is an eigenvalue of {$K$} with eigenvector {$v$}, then {$v=K^2v=\lambda^2v$}, thus {$\lambda =\pm 1$}.

This means that the vector space {$V=V_+\oplus V_-$} where {$Kv=v$} for {$v\in V_+$} and {$Kv=-v$} for {$v\in V_-$}.

Suppose {$A\in U(2n)$} and {$V_+$} and {$V_-$} are both n-dimensional. What can we conclude if {$iJ$} and {$A$} commute?

If {$v_+\in V_+$}, then {$iJAv_+=AiJv_+=Av_+$}. This implies {$Av_+\in V_+$}.

If {$v_-\in V_-$}, then {$iJAv_-=AiJv_-=-Av_-$}. This implies {$Av_-\in V_-$}.

Consequently, {$A\in U(n)\times U(n)$}. We have {$U(2n)\supset U(n)\times U(n)$}.

Similarly, consider {$K=J_1J_2J_3$}, where {$J_1J_2=-J_2J_1, J_1J_3=-J_3J_1, J_2J_3=-J_3J_2$}.

Then {$(J_1J_2J_3)(J_1J_2J_3) = J_1^2J_2J_3J_2J_3=-J_1^2J_2^2J_3^2=+1$}.

This yields {$O(2n)\supset O(n)\times O(n)$} and {$Sp(2n)\supset Sp(n)\times Sp(n)$}.

Note that {$(J_1J_2)^2=–1$}. For {$(J_1J_2)(J_1J_2)=-J_1^2J_2^2=-1$}. So this won't give us {$K$}.