The path of complex structure.

Question

$J = \left( {\begin{array}{*{20}{c}} 0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)$. If $I(t)$ is a path in $\rm M(2n, \mathbb R)$ ($2n \times 2n $ real matrix) such that $I(0)=0$ and $I(t)J+JI(t)=0$, then for sufficiently small $t$ can we find a path $J(t) \in \rm M(2n, \mathbb R)$ such that (1) $J(0)=J$, (2) $J^2(t)+I^2(t)=-I_{2n}$, (3) $I(t)J(t)+J(t)I(t)=0$?

Answer 1

Define the path $J(t)\in \rm M(2n, \mathbb R)$ as a matrix whos elements are smooth functions of $t$. This condition ensures the derivative of the matrix exists w.r.t. $t$ and thus we may Taylor expand about the point $t=0$. We then write:

$J(t)=J_0+J_1t+O(t^2)$: $J_n=\dfrac{1}{n!}\left(\dfrac{d}{dt}J(t)\right)|_{t=0}$,

$I(t)=I_0+I_1t+O(t^2)$: $I_n=\dfrac{1}{n!}\left(\dfrac{d}{dt}I(t)\right)|_{t=0}$,

where we have discarded matrices multiplied by powers of $t>1$ as we are required to demonstrate the existence of $J(t)$ for "sufficiently small $t$". Now we know that $ I(0)=0$ and $J(0)=J$, so we have:

$J(t)=J+J_1t$,

$I(t)=I_1t$.

Then, because we know that:

$I(t)J+JI(t)=0$

we have that:

$I_1J+JI_1=0$.

Now we are required to satisfy:

$J^2(t)+I^2(t)=-I_{2n}$,

which, with our first order expansions first becomes:

$J^2+J_1²t²+(JJ_1+J_1J)t+I_1²t²=-I_{2n}$

and then keeping up to first order in $t$:

$J^2+(JJ_1+J_1J)t=-I_{2n}$.

But:

$J² = \left( {\begin{array}{*{20}{c}} 0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)\left( {\begin{array}{*{20}{c}} 0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)=-I_{2n}$

and so this condition is met if:

$JJ_1+J_1J=0$.

Similarly, we are required to satisfy:

$I(t)J(t)+J(t)I(t)=0$

$I_1Jt+I_1J_1t²+JI_1t+J_1I_1t²=0$

$(I_1J+JI_1)t=0$ ... but we got this for free earlier.

So, we see that a path $J(t)$ can be found that meets all the requirements if we can find a $J_1$ that anti-commutes with $J$; that is if we can satisfy:

$JJ_1+J_1J=0$.

But we allready know one matrix that anti-commutes with $J$ and so we can solve this by letting:

$J_1=I_1$.

Thus the path:

$J(t)=J+I_1t$

satisfies the requirements; which is equivalent to saying that for sufficiently small $t$ the path:

$J(t)=J+I(t)$

satisfies the requirements. Finally, as $J,I(t)\in \rm M(2n, \mathbb R)$ it follows that $J(t)\in \rm M(2n, \mathbb R)$.

Answer 2

This is not what was asked for exactly, but the path exists in the complex domain...

Let the dot denote matrix multiplication to avoid function dependence ambiguity.

Take:

$J(t)=J.\left(\mathbb{I}_{2n}+iI(t)\right)\in \rm M(2n, \mathbb C): I(0)=0,J(0)=J$.

$J,I(t)\in \rm M(2n, \mathbb R)$

We have that:

$\{J(t),I(t)\}=J(t).I(t)+I(t).J(t)$

$=J.\left(\mathbb{I}_{2n}+iI(t)\right).I(t)+I(t).J.\left(\mathbb{I}_{2n}+iI(t)\right)$

$=\{J,I(t)\}.\left(\mathbb{I}_{2n}+iI(t)\right)$,

and so, because $\{J,I(t)\}=0$,

$\{J(t),I(t)\}=0$.

We also have that:

$J(t)²=J.\left(\mathbb{I}_{2n}+iI(t)\right).J.\left(\mathbb{I}_{2n}+iI(t)\right)$,

$J(t)²=J.J.\left(\mathbb{I}_{2n}-iI(t)\right).\left(\mathbb{I}_{2n}+iI(t)\right)$,

$J(t)²=J².\left(\mathbb{I}_{2n}+I(t)²\right)$,

but:

$J² = \left( {\begin{array}{*{20}{c}} 0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right).\left( {\begin{array}{*{20}{c}} 0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)=-\mathbb{I}_{2n}$,

and so:

$J(t)²+I(t)²=-\mathbb{I}_{2n}$.

Answer 3

Post-bounty remarks: I am not experienced in complex structures and the like, but here are some ideas I had before I offered a bounty to this question.

Note that, if we add together the equations \begin{align*} J^2(t)+I^2(t)&=-I_{2n}\tag{a},\\ I(t)J(t)+J(t)I(t)&=0,\tag{b} \end{align*} we get $$ (I(t)+J(t))^2=-I_{2n}.\tag{c} $$ In other words, the original problem can be transformed into the existence problem of a square root $R(t)=I(t)+J(t)$ of $-I_{2n}$ such that the initial condition $R(0)=J$ and the equivalent form of $\text{(b)}$, $$ I(t)R(t)+R(t)I(t) = 2I(t)^2,\tag{d} $$ are satisfied.

Answer 4

OK last go I promise...

$[A,B]=AB-BA$,

$\{ A,B \} =AB+BA$.

Define the following matrix under the assumption the power series converges :

$\sqrt{\mathbb{I}_{2n}+I(t)^2}=\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^k(2k)!}{(1-2k)(k!)^2(4^k)}I(t)^{2k}\in M(2n,\mathbb{R})$,

...I think it converges for "sufficiently small t", because if $I(t)\rightarrow0$ as $t\rightarrow0$ then the elements can be made arbitrarily small at some $t$; the equivalent 1d series converges for $I(t)<1$.

Note the following properties:

$\left[I(t),\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$,

because:

$\left[I(t),I(t)^{2n}\right]=0$,

and:

$\left[J,\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$,

because:

$\{J,I(t)\}=0 \Rightarrow J.I(t)^n=(-1)^nI(t)^n.J \Rightarrow [J,I(t)^{2n}]=0$.

With that said, let us define the following path:

$J(t)=J.\sqrt{\mathbb{I}_{2n}+I(t)^2}\in \rm M(2n, \mathbb{ R})$,

$I(0)=0,J(0)=J: J,I(t)\in \rm M(2n, \mathbb R)$.

Then note:

$\{J(t),I(t)\}=J.\sqrt{\mathbb{I}_{2n}+I(t)^2}.I(t)+I(t).J.\sqrt{\mathbb{I}_{2n}+I(t)^2}=\{J,I(t)\}.\sqrt{\mathbb{I}_{2n}+I(t)^2}=0$,

because:

$\left[I(t),\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$ ...and... $\{J,I(t)\}=0$,

and:

$J(t)^2+I(t)^2=J.\sqrt{\mathbb{I}_{2n}+I(t)^2}.J.\sqrt{\mathbb{I}_{2n}+I(t)^2}+I(t)^2=J^2.\left(\mathbb{I}_{2n}+I(t)^2\right)+I(t)^2=-\mathbb{I}_{2n}$,

because:

$\left[J,\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$ ...and... $J^2 = \left( {\begin{array}{*{20}{c}} 0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right).\left( {\begin{array}{*{20}{c}} 0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)=-\mathbb{I}_{2n}$.

NOTE: If we were forbidden to use the power series to prove the commutation relations, perhaps diagonalization may work. For instance if $I(t)$ is diaganolized by $S$ to leave a diagonal matrix $D$, with $D_{i,i}=e_i$ the eigenvalues of $S$, then we could define:

$I(t)=SDS^{-1},I(t)^2=SD^2S^{-1},\sqrt{\mathbb{I}_{2n}+I(t)^2}=S\sqrt{\mathbb{I}_{2n}+D^2}S^{-1}$,

this would certainly prove:

$\left[I(t),\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$,

but it’s not yet obvious how it might prove:

$\left[J,\sqrt{\mathbb{I}_{2n}+I(t)^2}\right]=0$.