Because of Euler’s identity , the complex exponential is not injective: for any complex and integer . As such, the complex logarithm is not well-defined as a single-valued function from to . However, after making a branch cut, one can create a branch of the logarithm which is single-valued. For instance, after removing the negative real axis , one has the standard branch of the logarithm, with defined as the unique choice of the complex logarithm of whose imaginary part has magnitude strictly less than . This particular branch has a number of useful additional properties:
- The standard branch is holomorphic on its domain .
- One has for all in the domain . In particular, if is real, then is real.
- One has for all in the domain .
One can then also use the standard branch of the logarithm to create standard branches of other multi-valued functions, for instance creating a standard branch of the square root function. We caution however that the identity can fail for the standard branch (or indeed for any branch of the logarithm).
One can extend this standard branch of the logarithm to complex matrices, or (equivalently) to linear transformations on an -dimensional complex vector space , provided that the spectrum of that matrix or transformation avoids the branch cut . Indeed, from the spectral theorem one can decompose any such as the direct sum of operators on the non-trivial generalised eigenspaces of , where ranges in the spectrum of . For each component of , we define
where is the Taylor expansion of at ; as is nilpotent, only finitely many terms in this Taylor expansion are required. The logarithm is then defined as the direct sum of the .
The matrix standard branch of the logarithm has many pleasant and easily verified properties (often inherited from their scalar counterparts), whenever has no spectrum in :
- (i) We have .
- (ii) If and have no spectrum in , then .
- (iii) If has spectrum in a closed disk in , then , where is the Taylor series of around (which is absolutely convergent in ).
- (iv) depends holomorphically on . (Easily established from (ii), (iii), after covering the spectrum of by disjoint disks; alternatively, one can use the Cauchy integral representation for a contour in the domain enclosing the spectrum of .) In particular, the standard branch of the matrix logarithm is smooth.
- (v) If is any invertible linear or antilinear map, then . In particular, the standard branch of the logarithm commutes with matrix conjugations; and if is real with respect to a complex conjugation operation on (that is to say, an antilinear involution), then is real also.
- (vi) If denotes the transpose of (with the complex dual of ), then . Similarly, if denotes the adjoint of (with the complex conjugate of , i.e. with the conjugated multiplication map ), then .
- (vii) One has .
- (viii) If denotes the spectrum of , then .
As a quick application of the standard branch of the matrix logarithm, we have
Proposition 1 Let be one of the following matrix groups: , , , , , or , where is a non-degenerate real quadratic form (so is isomorphic to a (possibly indefinite) orthogonal group for some . Then any element of whose spectrum avoids is exponential, that is to say for some in the Lie algebra of .
Proof: We just prove this for , as the other cases are similar (or a bit simpler). If , then (viewing as a complex-linear map on , and using the complex bilinear form associated to to identify with its complex dual , then is real and . By the properties (v), (vi), (vii) of the standard branch of the matrix logarithm, we conclude that is real and , and so lies in the Lie algebra , and the claim now follows from (i).
Exercise 2 Show that is not exponential in if . Thus we see that the branch cut in the above proposition is largely necessary. See this paper of Djokovic for a more complete description of the image of the exponential map in classical groups, as well as this previous blog post for some more discussion of the surjectivity (or lack thereof) of the exponential map in Lie groups.
For a slightly less quick application of the standard branch, we have the following result (recently worked out in the answers to this MathOverflow question):
Proposition 3 Let be an element of the split orthogonal group which lies in the connected component of the identity. Then .
The requirement that lie in the identity component is necessary, as the counterexample for shows.
Proof: We think of as a (real) linear transformation on , and write for the quadratic form associated to , so that . We can split , where is the sum of all the generalised eigenspaces corresponding to eigenvalues in , and is the sum of all the remaining eigenspaces. Since and are real, are real (i.e. complex-conjugation invariant) also. For , the restriction of to then lies in , where is the restriction of to , and
The spectrum of consists of positive reals, as well as complex pairs (with equal multiplicity), so . From the preceding proposition we have for some ; this will be important later.
It remains to show that . If has spectrum at then we are done, so we may assume that has spectrum only at (being invertible, has no spectrum at ). We split , where correspond to the portions of the spectrum in , ; these are real, -invariant spaces. We observe that if are generalised eigenspaces of with , then are orthogonal with respect to the (complex-bilinear) inner product associated with ; this is easiest to see first for the actual eigenspaces (since for all ), and the extension to generalised eigenvectors then follows from a routine induction. From this we see that is orthogonal to , and and are null spaces, which by the non-degeneracy of (and hence of the restriction of to ) forces to have the same dimension as , indeed now gives an identification of with . If we let be the restrictions of to , we thus identify with , since lies in ; in particular is invertible. Thus
and so it suffices to show that .
At this point we need to use the hypothesis that lies in the identity component of . This implies (by a continuity argument) that the restriction of to any maximal-dimensional positive subspace has positive determinant (since such a restriction cannot be singular, as this would mean that positive norm vector would map to a non-positive norm vector). Now, as have equal dimension, has a balanced signature, so does also. Since , already lies in the identity component of , and so has positive determinant on any maximal-dimensional positive subspace of . We conclude that has positive determinant on any maximal-dimensional positive subspace of .
We choose a complex basis of , to identify with , which has already been identified with . (In coordinates, are now both of the form , and for .) Then becomes a maximal positive subspace of , and the restriction of to this subspace is conjugate to , so that
But since and is positive definite, so as required.