## The Golden-Thompson inequality

^{Posted at Fri, 27 Aug 2010 13:48:17 GMT}

Let be two Hermitian matrices. When and commute, we have the identity

When and do not commute, the situation is more complicated; we have the

where the infinite product here is explicit but very messy. On the other hand, taking determinants we still have the identity

Recently I learned (from Emmanuel Candes, who in turn learned it from David Gross) that there is another very nice relationship between and , namely the Golden-Thompson inequality

The remarkable thing about this inequality is that no commutativity hypotheses whatsoever on the matrices are required. Note that the right-hand side can be rearranged using the cyclic property of trace as ; the expression inside the trace is positive definite so the right-hand side is positive. (On the other hand, there is no reason why expressions such as need to be positive or even real, so the obvious extension of the Golden-Thompson inequality to three or more Hermitian matrices fails.) I am told that this inequality is quite useful in statistical mechanics, although I do not know the details of this.

To get a sense of how delicate the Golden-Thompson inequality is, let us expand both sides to fourth order in . The left-hand side expands as

while the right-hand side expands as

Using the cyclic property of trace , one can verify that all terms up to third order agree. Turning to the fourth order terms, one sees after expanding out and using the cyclic property of trace as much as possible, we see that the fourth order terms

Intuitively, the Golden-Thompson inequality is asserting that interactions between a pair of non-commuting Hermitian matrices are strongest when cross-interactions are kept to a minimum, so that all the factors lie on one side of a product and all the factors lie on the other. Indeed, this theme will be running through the proof of this inequality, to which we now turn.

The proof of the Golden-Thompson inequality relies on the somewhat magical power of the tensor power trick. For any even integer and any matrix (not necessarily Hermitian), we define the -Schatten norm of by the formula

(This formula in fact defines a norm for any , but we will only need the even integer case here.) This norm can be viewed as a non-commutative analogue of the norm; indeed, the -Schatten norm of a

Note that the -Schatten norm

is the Hilbert space norm associated to the Frobenius inner product (or Hilbert-Schmidt inner product)

This is clearly a non-negative Hermitian inner product, so by the Cauchy-Schwarz inequality we conclude that

for any matrices . As , we conclude in particular that

We can iterate this and establish the

whenever is an even power of . Indeed, we induct on , the case already having been established. If is a power of , then by the induction hypothesis (grouping into pairs) we can bound

On the other hand, we may expand

We use the cyclic property of trace to move the rightmost factor to the left. Applying the induction hypothesis again, we conclude that

But from the cyclic property of trace again, we have and . We conclude that

and similarly for , etc. Inserting this into (3) we obtain (2).

Specialising (2) to the case where for some Hermitian matrices , we conclude that

and hence by cyclic permutation

for any . Iterating this we conclude that

Applying this with replaced by and respectively, we obtain

Now we send . Since and , we have , and so the left-hand side is ; taking the limit as we obtain the Golden-Thompson inequality. (See also these notes of Vershynin for a slight variant of this proof.)

If we stop the iteration at an earlier point, then the same argument gives the inequality

for a power of two; one can view the original Golden-Thompson inequality as the endpoint of this case in some sense. (In fact, the Golden-Thompson inequality is true in any operator norm; see Theorem 9.3.7 of Bhatia

When and do not commute, the situation is more complicated; we have the

*Baker-Campbell-Hausdorff formula*where the infinite product here is explicit but very messy. On the other hand, taking determinants we still have the identity

Recently I learned (from Emmanuel Candes, who in turn learned it from David Gross) that there is another very nice relationship between and , namely the Golden-Thompson inequality

The remarkable thing about this inequality is that no commutativity hypotheses whatsoever on the matrices are required. Note that the right-hand side can be rearranged using the cyclic property of trace as ; the expression inside the trace is positive definite so the right-hand side is positive. (On the other hand, there is no reason why expressions such as need to be positive or even real, so the obvious extension of the Golden-Thompson inequality to three or more Hermitian matrices fails.) I am told that this inequality is quite useful in statistical mechanics, although I do not know the details of this.

To get a sense of how delicate the Golden-Thompson inequality is, let us expand both sides to fourth order in . The left-hand side expands as

while the right-hand side expands as

Using the cyclic property of trace , one can verify that all terms up to third order agree. Turning to the fourth order terms, one sees after expanding out and using the cyclic property of trace as much as possible, we see that the fourth order terms

*almost*agree, but the left-hand side contains a term whose counterpart on the right-hand side is . The difference between the two can be factorised (again using the cyclic property of trace) as . Since is skew-Hermitian, is positive definite, and so we have proven the Golden-Thompson inequality to fourth order. (One could also have used the Cauchy-Schwarz inequality for the Frobenius norm to establish this; see below.)Intuitively, the Golden-Thompson inequality is asserting that interactions between a pair of non-commuting Hermitian matrices are strongest when cross-interactions are kept to a minimum, so that all the factors lie on one side of a product and all the factors lie on the other. Indeed, this theme will be running through the proof of this inequality, to which we now turn.

The proof of the Golden-Thompson inequality relies on the somewhat magical power of the tensor power trick. For any even integer and any matrix (not necessarily Hermitian), we define the -Schatten norm of by the formula

(This formula in fact defines a norm for any , but we will only need the even integer case here.) This norm can be viewed as a non-commutative analogue of the norm; indeed, the -Schatten norm of a

*diagonal*matrix is just the norm of the coefficients.Note that the -Schatten norm

is the Hilbert space norm associated to the Frobenius inner product (or Hilbert-Schmidt inner product)

This is clearly a non-negative Hermitian inner product, so by the Cauchy-Schwarz inequality we conclude that

for any matrices . As , we conclude in particular that

We can iterate this and establish the

*non-commutative H?lder inequality*whenever is an even power of . Indeed, we induct on , the case already having been established. If is a power of , then by the induction hypothesis (grouping into pairs) we can bound

On the other hand, we may expand

We use the cyclic property of trace to move the rightmost factor to the left. Applying the induction hypothesis again, we conclude that

But from the cyclic property of trace again, we have and . We conclude that

and similarly for , etc. Inserting this into (3) we obtain (2).

Remark 1Though we will not need to do so here, it is interesting to note that one can use the tensor power trick to amplify (2) for equal to a power of two, to obtain (2) for all positive integers , at least when the are all Hermitian. Indeed, pick a large integer and let be the integer part of . Then expand the left-hand side of (2) as and apply (2) with replaced by to bound this by . Sending (noting that ) we obtain the claim.

Specialising (2) to the case where for some Hermitian matrices , we conclude that

and hence by cyclic permutation

for any . Iterating this we conclude that

Applying this with replaced by and respectively, we obtain

Now we send . Since and , we have , and so the left-hand side is ; taking the limit as we obtain the Golden-Thompson inequality. (See also these notes of Vershynin for a slight variant of this proof.)

If we stop the iteration at an earlier point, then the same argument gives the inequality

for a power of two; one can view the original Golden-Thompson inequality as the endpoint of this case in some sense. (In fact, the Golden-Thompson inequality is true in any operator norm; see Theorem 9.3.7 of Bhatia