UDC 517.941.92

REFINEMENT OF A THEOREM OF V. V. MOROZOV

A. G. ASLANYAN, V. I. BURENKOV

1. We consider a system of linear differential equations

\[ \frac{dX}{dt}=X[U_1\varphi_1(t)+U_2\varphi_2(t)], \tag{1} \]

where \(U_1\) and \(U_2\) are constant square matrices of order \(n\), generally complex; \(\varphi_1(t)\) and \(\varphi_2(t)\) are continuous scalar functions of \(t\); \(X\) is an unknown matrix.

N. P. Erugin [1] posed the following problem: to find conditions that the matrices \(U_1\) and \(U_2\) must satisfy so that the integral matrix \(X(t)\), normalized at the point \(t=t_0\), can, for arbitrary continuous \(\varphi_1(t)\) and \(\varphi_2(t)\), be represented in the form

\[ X=e^S e^{U_2\psi_2}, \tag{2} \]

where \(\psi_2(t)=\displaystyle\int_{t_0}^{t}\varphi_2(\tau)\,d\tau\), and the matrix \(S\) commutes with its derivative, i.e.

\[ S\frac{dS}{dt}=\frac{dS}{dt}S. \]

Substituting (2) into equation (1), we obtain a differential equation for \(S\), whence

\[ S=\int_{t_0}^{t} e^{U_2\psi_2}U_1e^{-U_2\psi_2}\varphi_1\,dt, \tag{3} \]

i.e. in this case the solution of system (1) is found by quadratures.

V. V. Morozov [2] proved the following theorem.

In order that the problem of N. P. Erugin have a solution, it is necessary and sufficient that the matrices \(U_1\) and \(U_2\) satisfy the following infinite nonlinear system of equations:

\[ \mathfrak{A}_1=[U_1[U_2U_1]]=0, \]

\[ \mathfrak{A}_2=[U_1[U_2[U_2U_1]]]=0, \]

\[ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot \tag{4} \]

\[ \mathfrak{A}_k=[U_1[U_2\ldots[U_2U_1]\ldots]]=0, \]

\[ \underbrace{\hphantom{[U_2\ldots[U_2U_1]\ldots]}}_{k+1\ \text{brackets}} \]

\[ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot \]

where \([AB]=AB-BA\).

2. In the present paper we shall prove that the equations of system (4) are not independent; namely, the equations \(\mathfrak A_{2k}=0\) \((k=1,2,\ldots)\) are consequences of the equations \(\mathfrak A_{2l+1}=0\) \((l=0,1,2,\ldots)\). More precisely, we shall establish the following result.

Denote
\[ [A^nB]=[A[A\ldots[AB]\ldots]] \]
(\(n\) brackets).

Theorem 1. The following formula holds:
\[ \mathfrak A_{2k}=\sum_{l=1}^{k}\alpha_l^{(k)} [U_2^{\prime\,2(k-l)+1}\mathfrak A_{2l-1}], \tag{5} \]
where the numbers \(\alpha_l^{(k)}\) satisfy the triangular system of equations
\[ \sum_{l=i+1}^{k} C_{2l-1}^{2i}\alpha_l^{(k)} = C_{2k}^{2i} \qquad (i=0,1,\ldots,k-1). \tag{6} \]

It is clear from formula (5) that if \(\mathfrak A_1=0,\ \mathfrak A_3=0,\ldots,\mathfrak A_{2k-1}=0\), then also \(\mathfrak A_{2k}=0\). Combining the result of V. V. Morozov formulated above with Theorem 1, we arrive at the following theorem on the solvability of equation (1) in the form (2).

Theorem 2. In order that the problem of N. P. Erugin have a solution, it is necessary and sufficient that the matrices \(U_1\) and \(U_2\) satisfy the following system of equations:
\[ \mathfrak A_{2k+1}=0\qquad (k=0,1,2,\ldots). \tag{7} \]

3. Proof of Theorem 1. The plan of the proof is as follows. We shall use the fact that the \(\mathfrak A_k\) belong to the free Lie ring \(\Lambda\) generated by the elements \(U_1\) and \(U_2\). We define inductively (see [3]) the standard commutators of the free Lie ring \(\Lambda\) in the following way:

1. \(U_1,\ U_2\) are standard commutators of degree 1. Suppose that we have defined standard commutators of degrees \(1,2,\ldots,n-1\); order them so that \(U<V\) whenever the degree of \(U\) is less than the degree of \(V\).

2. If the degree of \(U\) is \(r\), the degree of \(V\) is \(s\), and \(r+s=n\), then \([U,V]\) is a standard commutator of degree \(n\) if
   a) \(U\) and \(V\) are standard commutators and \(U>V\);
   b) and, moreover, if the standard commutator \(U\) has the form \([T,W]\), then \(V\geq W\).

M. Hall [3] proved the following theorem (generally speaking, for the general case when the free Lie ring \(\Lambda\) is generated by the elements \(U_1,U_2,\ldots,U_n\)):

The standard commutators form a basis in the free Lie ring \(\Lambda\) generated by the elements \(U_1\) and \(U_2\).

Let us single out in the free Lie ring \(\Lambda\) the subset of all elements having degree \(2k+2\) (\(k\geq 1\) is fixed) and such that the degree of the element \(U_1\) is equal to 2. Denote this subset by \(M_{2k+2}\). Obviously, \(M_{2k+2}\) is a linear subspace. Using the basis of standard commutators, we shall prove that the dimension of this subspace is
\[ \dim M_{2k+2}=k \]
(see § 3.1).

Next, we shall prove (see § 3.2) that the \(k\) elements of the form
\[ [U_2^{\prime\,2(k-l)+1}\mathfrak A_{2l-1}] \qquad (l=1,2,\ldots,k), \]
which have degree \(2k+2\), are linearly independent and, consequently, form...

a basis in the subspace \(M_{2k+2}\). Since the element \(\mathfrak A_{2k}\in M_{2k+2}\), it is (uniquely) expressed in terms of the basis elements, i.e.

\[ \mathfrak A_{2k}=\sum_{l=1}^{k}\alpha_l^{(k)} \bigl[U_2^{\,2(k-l)+1}\mathfrak A_{2l-1}\bigr], \]

which proves formula (5).

Finally, one can show (see Sec. 3.3) that the numbers \(\alpha_l^{(k)}\) satisfy system (6).

3.1. \(\dim M_{2n}=n-1\).

In what follows we shall everywhere assume that \(U_1<U_2\). Let us first consider the subset of elements of degree \(k\) in the free Lie ring \(\Lambda\) that contain the element \(U_1\) in the first degree. Denote this subset by \(L_k\). We shall prove that \(L_k\), for arbitrary \(k=1,2,\ldots\), contains only one standard commutator

\[ \omega_k= \begin{cases} U_1, & \text{for } k=1,\\[4pt] [\ldots [U_2U_1]U_2]\ldots U_2], \quad \underbrace{\phantom{[\ldots [U_2U_1]U_2]\ldots U_2]}}_{k-1\ \text{brackets}} & \text{for } k\ge 2, \end{cases} \]

i.e. \(\dim L_k=1\).

Indeed, any standard commutator belonging to \(L_k\) must be formed from two standard commutators, one of which must contain the element \(U_1\) in the first degree, while the other must not contain the element \(U_1\) at all. But there is only one standard commutator not containing the element \(U_1\), namely the element \(U_2\). Hence a standard commutator containing the element \(U_1\) in the first degree has degree \(k-1\) and belongs to \(L_{k-1}\). If we now assume that the subset \(L_{k-1}\) contains only one standard commutator

\[ [\ldots [U_2U_1]U_2]\ldots U_2], \quad \underbrace{\phantom{[\ldots [U_2U_1]U_2]\ldots U_2]}}_{k-2\ \text{brackets}} \]

then it follows from the preceding reasoning that the subset \(L_k\) also contains one standard commutator

\[ [[\ldots [U_2U_1]U_2]\ldots U_2]U_2]. \quad \underbrace{\phantom{[[\ldots [U_2U_1]U_2]\ldots U_2]U_2]}}_{k-1\ \text{brackets}} \]

Since our assertion is obvious for \(L_1\) and \(L_2\), it follows in the general case as well that \(\dim L_k=1\).

Let us now consider the subset \(M_{2n}\), \(n>1\). A standard commutator from \(M_{2n}\) can be formed in two ways: 1) from two standard commutators \(\omega_i\) and \(\omega_j\), with \(i+j=n\); 2) from a standard commutator \(z\in M_{2n-1}\) and the standard commutator \(U_2\).

In the first case one can formally form \(n-1\) pairs

\[ [\omega_{2n-1},\omega_1],\quad [\omega_{2n-2},\omega_2],\ \ldots,\ [\omega_{n+1},\omega_{n-1}], \]

satisfying condition 2a), but the commutator \([\omega_{2n-1},\omega_1]\) is not a standard commutator, since

\[ \omega_1=U_1,\qquad \omega_{2n-1}= [\ldots [U_2U_1]\ldots U_2]\equiv [t,U_2] \quad \underbrace{\phantom{[\ldots [U_2U_1]\ldots U_2]}}_{2n-2\ \text{brackets}} \]

and \(U_1<U_2\), i.e. condition 2b) is not fulfilled. It is also obvious that all the remaining commutators

\[ [\omega_{2n-2},\omega_2],\ \ldots,\ [\omega_{n+1},\omega_{n-1}] \]

are standard and, consequently, by the first method one can form \(n-2\) standard commutators belonging to \(M_{2n}\).

Let us consider the second method of forming standard commutators belonging to \(M_{2n}\). In this case the standard commutator has the form \(t=[z,U_2]\), where \(z\in M_{2n-1}\). Further, \(z=[z_1V_1]\), where \(V_1=U_1\) or \(V_1=U_2\) (since, according to condition 2b), \(U_2>V_1\)). Similarly,

\[ z_1=[z_2,V_2],\qquad \text{where } V_2=U_1 \text{ or } V_2=U_2; \]

\[ z_2=[z_3,V_3],\qquad \text{where } V_3=U_1 \text{ or } V_3=U_2, \]

and so on.

We shall show that if \(V_j=U_1\), then the degree of the commutator \(t\) is equal to \(j+3\). Indeed, then \(z_j=[z_{j+1},V_{j+1}]\) and, by condition 2b), \(V_j>V_{j+1}\), i.e. \(U_1>V_{j+1}\), whence \(V_{j+1}=U_1\). But the commutator \(t\) contains the element \(U_1\) only in the second degree, whence it follows that \(z_{j+1}=U_2\) and the degree of \(t\) is \(j+3\). Since \(t\in M_{2n}\), we have \(j+3=2n\), or \(j=2n-3\), and there is a unique standard commutator

\[ z=\underbrace{[\ldots [[U_2U_1]U_1]U_2]\ldots U_2]}_{2n-2\ \text{brackets}}, \]

belonging to \(M_{2n-1}\), by means of which, together with the element \(U_2\), one can form the standard commutator \(t\in M_{2n}\), \(t=[z,U_2]\). Combining the results obtained, we see that

\[ \dim M_{2n}=n-1. \]

3.2. The elements

\[ [U_2^{\{2(k-i)+1\}}\mathfrak A_{2i-1}] \qquad (i=1,\ldots,k) \tag{8} \]

are linearly independent.

By induction, using the relation \(C_k^i+C_k^{i-1}=C_{k+1}^i\), it is not hard to establish that

\[ [A^{\{n\}}B]=\sum_{i=0}^{n}(-1)^i C_n^i A^{\,n-i}BA^i. \tag{9} \]

Let us now equate to zero a linear combination of elements of the form (8):

\[ \sum_{i=1}^{k}\beta_i[U_2^{\{2(k-i)+1\}}\mathfrak A_{2i-1}]=0. \tag{10} \]

By formula (9) we have

\[ \mathfrak A_m=[U_1[U_2^{\{m\}}U_1]] =U_1[U_2^{\{m\}}U_1]-[U_2^{\{m\}}U_1]U_1= \]

\[ =\sum_{s=0}^{m}(-1)^s C_m^s U_1U_2^{\,m-s}U_1U_2^s -\sum_{s=0}^{m}(-1)^s C_m^s U_2^{\,m-s}U_1U_2^sU_1, \tag{11} \]

and equality (10) will take the form

\[ \sum_{i=1}^{k}\beta_i \sum_{j=0}^{2(k-i)+1} \sum_{s=0}^{2i-1} (-1)^{j+s}C_{2(k-i)+1}^{j}C_{2i-1}^{s}\times \]

\[ \times\{\,U_2^{\,2(k-i)+1-j}U_1U_2^{\,2i-1-s}U_1U_2^{\,s+j} -U_2^{\,2k-j-s}U_1U_2^sU_1U_2^j\,\}=0. \tag{12} \]

If we now set equal to zero the coefficients of \(U_1U_2^{2l}U_1U_2^{2k-2l}\) \((l=0,1,\ldots,k-1)\), then for determining \(\beta_i\) we obtain a homogeneous triangular system of equations

\[ \sum_{i=l+1}^{k} C_{2i-1}^{2l}\beta_i=0 \qquad (l=0,1,\ldots,k-1). \]

Since the determinant of this system is not equal to zero, it follows that

\[ \beta_1=\cdots=\beta_k=0, \]

whence it follows that the elements (8) are linearly independent.

3.3. The numbers \(a_i^{(k)}\) satisfy the system of equations

\[ \sum_{i=l+1}^{k} C_{2i-1}^{2l}a_i^{(k)}=C_{2k}^{2l} \qquad (l=0,1,\ldots,k-1). \tag{6'} \]

Expanding the left- and right-hand sides of formula (5), we obtain an equality whose left-hand side differs from the left-hand side of (12) only in that \(a_i^{(k)}\) stands there instead of \(\beta_i\), while on the right stands expression (11) for \(m=2k\). Equating the coefficients of

\[ U_1U_2^{2l}U_1U_2^{2k-2l} \qquad (l=0,1,\ldots,k-1), \]

we obtain the desired system for \(a_i^{(k)}\).

References

1. Erugin N. P. Izv. AN SSSR, Mathematics, 5, 377—380, 1941.
2. Morozov V. V. Izv. vuzov, Mathematics, No. 5, 171—173, 1959.
3. Hall M. Proceedings of the American Math. Soc., 1, No. 5, 575—581, 1950.

Received by the editors
December 6, 1965

Moscow Institute of Physics and Technology