Whereas the conventional "sum integral" is $$ \lim_{\Delta x\to 0} \sum_i f(x_i)\,\Delta x, $$ a "product integral" is $$ \lim_{\Delta x\to 0} \prod_i f(x_i)^{\Delta x}. $$

Now you're thinking: just take logarithms and it's a "sum integral", so why bother having this additional concept? Somewhere I heard this answer proposed: Because one does this with matrix multiplication rather than multiplication of numbers.


  • Is this actually done? What's been published on it?
  • What are the most interesting results about it?
  • What's it's used for?
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    $\begingroup$ I think this was written by Terry Tao here: cornellmath.wordpress.com/2008/01/26/… $\endgroup$ Aug 15, 2013 at 20:12
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    $\begingroup$ In high-energy and condensed matter physics, the representation of the propagator/Green's function in Feynman's path integral formulation is essentially an infinite "product integral" of matrices. Since a good chunk of quantum mechanics/field theory can be formulated on top of "path integral", you can say nearly all quantum physics has good uses of this concept. $\endgroup$ Aug 15, 2013 at 20:35
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    $\begingroup$ @achillehui : Maybe you could expand a bit and make this into an answer. $\endgroup$ Aug 15, 2013 at 21:49
  • $\begingroup$ @deoxygerbe : Where in Terry Tao's posting do you see anything about products of matrices? He's writing about infinite products, and the question above is not about infinite products. $\endgroup$ Nov 26, 2017 at 15:50
  • $\begingroup$ @MichaelHardy: "The existence of the logarithm function does make the theory of infinite products of scalars essentially equivalent to the theory of infinite series, but the subject becomes significantly richer when one works with infinite products of matrices or operator", third comment in the link? $\endgroup$ Nov 30, 2017 at 14:32

1 Answer 1


The operation you are looking for is very useful in physics, probability, and geometry. In physics it goes under the name of "ordered exponential", but it is exactly the same: http://en.wikipedia.org/wiki/Ordered_exponential. Its series expansion is called "Dyson series", and it approximates the same thing: http://en.wikipedia.org/wiki/Dyson_series

In differential geometry, it gives the parallel transport map from a connection (the parallel transport map is the "line product integral" of the connection form).

All the uses of such an operation can be traced back to the solution of the following equation: $$ \frac{dX}{dt} = A(t)X(t), $$

where $X$ is either a vector or a (group) matrix, and $A$ is a (Lie algebra) matrix. Schrödinger's equation in quantum physics has this form, the parallel transport equation in differential geometry has this form (vanishing covariant derivative), and many more. The solution is: $$ X(t) = \prod_0^t \exp \big( A(t')\,dt' \big)\,X(0). $$

A very simple example is the following. If we have a rotation in the plane, we can define the angular velocity as: $$ \omega := \frac{d\theta}{dt}. $$

Therefore: $$ \theta = \int_0^t \omega (t')\,dt' +\theta(0). $$

If the rotation takes place in three dimension, this picture breaks down, because there is no "number" parametrizing the rotation angle. We have a rotation matrix, $R\in SO(3)$, and its derivative is an element of the Lie algebra $\Omega \in so(3)$. We have (can you see why?): $$ R(t) = \prod_0^t \exp \big( \Omega(t')\,dt' \big). $$

In the case of rotations in the plane everything was simple because $SO(2)$ is commutative, so the product integral, there, is exactly the exponential of the integral (just like with numbers).

Intuitively: if instead of "continuously summing" you need to "continuously compose group operations", you need a product integral. You can use an ordinary integral (and exponentiation) if and only if the group is commutative.


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