Is it possible to make ncalgebra factorize commutative coefficients to the left?

For example, if I define a non-commutative object W and a commutative object a by


and then write

a W

the output is, as expected, just a W. Same for

W a

However, if I write

a[t] W

instead, for some reason a[t] lands on the right and the output is

W a[t]

Even if I apply Simplify to the latter, it still remains the same.

Finally, if I write

a[t] W + a[t] tp[W] // Simplify

then the output looks as I would want it to:

a[t] (W + tp[W])

I found two similar questions on the forum, but the approach with NCCollect doesn't work here.

Here's a more real life example: enter image description here The commutative expression in a in the second parentheses got factored out to the left, but not the expression in the first.

By the way, the LeafCount of the expression factored as I want it to be is one less than the LeafCount of the output.

P.S. I found the following approach to my real life example: enter image description here First, it feels more like a kludge than a solution, but it does the job. Secondly, the commutative factor is still not on the left, but at least the formula it's taken out.

Is there a better way to do it?


1 Answer 1


I am not sure about what you want to do here but be careful. Even though you set a to be commutative, a[t] is still non commutative. This is because t, which is a small letter is still non commutative. If you set a and t to be commutative then a[t] would be treated as commutative. The way you have it now you have a commutative product (Times) of two non commutative expressions a[t] and W, and that is bound to cause you trouble, such as things don't factor as you would expect.

Regarding the ordering in a[t] W, this is controller by Times, which is commutative and sorted. There is nothing you can do about that.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.