Do all algebras have derivations?

[ajb]

A seemingly simple question, do all algebras have at least one non-trivial derivation?

 

Noncommutative algebras (lets say over complex or real numbers) always have at least the inner derivations. That is one can pick an element not in the centre and use the commutator to define a derivation. There may be outer derivations also*.

 

Now for the commutative case the same procedure will produce an operator that annihilates all elements in the algebra. So we need outer derivations. If the algebra is generated by a finite number of generators then one need only state the action on the generators and extend this to the whole algebra. So for polynomial algebras with the simple relation xy=yx this seems fine. One can then complete the algebra and define smooth functions, and we are essentially doing calculus as we all know it.

 

But what about more general commutative algebras lets say finite as a vector space, over the real or complex numbers with a unit? Is it known that all algebras have derivations or not?

 

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*Theorem: Every derivation of a finite dimensional centre simple algebra is inner.

 

So there is not always outer derivations.

Edited June 10, 2011 by ajb

[DrRocket]

A seemingly simple question, do all algebras have at least one non-trivial derivation?

 

Noncommutative algebras (lets say over complex or real numbers) always have at least the inner derivations. That is one can pick an element not in the centre and use the commutator to define a derivation. There may be outer derivations also*.

 

Now for the commutative case the same procedure will produce an operator that annihilates all elements in the algebra. So we need outer derivations. If the algebra is generated by a finite number of generators then one need only state the action on the generators and extend this to the whole algebra. So for polynomial algebras with the simple relation xy=yx this seems fine. One can then complete the algebra and define smooth functions, and we are essentially doing calculus as we all know it.

 

But what about more general commutative algebras lets say finite as a vector space, over the real or complex numbers with a unit? Is it known that all algebras have derivations or not?

 

-----------------------

 

*Theorem: Every derivation of a finite dimensional centre simple algebra is inner.

 

So there is not always outer derivations.

 

 

So far as I know an inner derivation is one that arises, as you describe frrom the commutator of some element in the algebra (or the adjoint representation in the case of Lie algebras, but I assume we are talking about associative algebras here).

 

I presume that an outer derivation is a derivation that is not inner, but have not seen the term defined.

 

That said, I do not have answer to your question. Maybe this paper could serve as a place to start looking, but I must admit to finding it outside of my interests and have not read it in any detail.

 

The question as stated is purely algebraic. Is there some motivation in analysis or geometry, or perhaps a more restrictive setting ? In settings with which I am more familiar derivations arise naturally, and one is not usually seeking just an arbitrary abstract derivation, but rather implications of the specific derivation or class of derivations that has presented itself (as in the identification with the tangent space at a point as the derivations on the germs of smooth functions).

[ajb]

I am thinking about associative algebras (maybe graded also). The question is not really well motivated other than the situation is clearer in the noncommutative setting. I have no idea what is generally known in the commutative case. Maybe it is not a very well founded question as I pose it.

 

The notion is of course far more general than the standard notion from real calculus, so one may have "vector fields" in a much more general setting. For noncommutative algebras we have some nice theorems in the case that they are centre simple and finite.

 

Really I am just "think out loud" here and wondered if anyone had any references close to my question. I will look up your suggestion.

[DrRocket]

I am thinking about associative algebras (maybe graded also). The question is not really well motivated other than the situation is clearer in the noncommutative setting. I have no idea what is generally known in the commutative case. Maybe it is not a very well founded question as I pose it.

 

The notion is of course far more general than the standard notion from real calculus, so one may have "vector fields" in a much more general setting. For noncommutative algebras we have some nice theorems in the case that they are centre simple and finite.

 

Really I am just "think out loud" here and wondered if anyone had any references close to my question. I will look up your suggestion.

 

 

There is a fairly long section on derivations in volume 1 of Zariski and Samuel (Commutative Algebra), but I didn't see anything pertinent to your question on a quick skim.

 

Until your post I had not run across the term "outer derivation". I presume it is just a derivation that does not arise from a commutator, hence in the abelian case any non-trivial derivation would be "outer".

 

I also assume that "centre simple" is British for "central simple". "Two people separated by a common language" -- George S, Patton

 

BTW "A bee in your bonnet" has nothing to do with an insect in your automobile..

Edited June 10, 2011 by DrRocket

[imatfaal]

"Two people separated by a common language" Seems we can't agree on anything not even who said that - first handful of Google searches give George Bernard Shaw, George S Patton, Oscar Wilde, Mark Twain, and Winston Churchill (2 yanks, 2 irish and 1 half American-English)

[DrRocket]

"Two people separated by a common language" Seems we can't agree on anything not even who said that - first handful of Google searches give George Bernard Shaw, George S Patton, Oscar Wilde, Mark Twain, and Winston Churchill (2 yanks, 2 irish and 1 half American-English)

 

Patton might have been quoting Shaw. George C. Scott uttered the phrase in the movie "Patton".

[ajb]

Maybe I meant central simple, but I am sure I have also read centre simple somewhere. It has not caused any real confusion here.

 

So an inner derivation of a noncommutative algebra is a derivation that is given by [math]X = [a, \bullet][/math] for some element [math]a[/math] in the algebra. It makes sense to consider elements not in the centre, but I don't think that matters too much. A derivation is known as outer if it cannot be written as a commutator.

 

The same definition holds for Lie algebras, but notice that the commutator between elements of a noncommutative algebra defines a Lie bracket so I doubt there is any real differences here at this level. Of course for noncommutative algebras the commutator gives us a Poisson bracket.

 

I take it there is not some "universal method" of producing derivations over a commutative algebra? Though, like I said for finitely generated polynomial algebras (with the commutation relation) and their related formal power series algebras the "standard derivatives" are derivations. These are probably the most complicated things of general interest, one can think of analytic and smooth manifolds in these terms.

 

I am not going to worry about this too much.

Edited June 12, 2011 by ajb

[khaled]

I've read that: some Lie Algebra can be with no outer derivation, and considered unsolvable

 

.. I wish you don't ask me any details, I'm not good in math

[uncool]

Clearly, the reals themselves don't have a derivation, as derivations are supposed to be linear, so D(x) = kx for some k, and D(ab) = kab, but aD(b) = kab, D(a)b = kab, so the right side is 2kab. The two are not equal unless k = 0, which is trivial.

=Uncool-

[ajb]

Clearly, the reals themselves don't have a derivation, as derivations are supposed to be linear, so D(x) = kx for some k,

 

Why does that have to be the same k for all x?

 

Linearity implies that [math]\delta(a+b) = \delta(a)+ \delta(b) [/math] for a differential field. Which is again hard to understand in your example.

Edited July 25, 2011 by ajb

[uncool]

Why does that have to be the same k for all x?

 

Linearity implies that [math]\delta(a+b) = \delta(a)+ \delta(b) [/math] for a differential field. Which is again hard to understand in your example.

You talked about algebras over the reals or complexes, so I assumed the derivations have to be linear over the reals. And linear over the reals means that for any elements x and y of the algebra, and reals a and b, [math]\delta(ax + by) = a\delta(x) + b\delta(y)[/math], not only what you said.

 

If we assume that the algebra is the reals, then let [math]k = \delta(1)[/math]. Then [math]\delta(x) = x \delta(1) = kx[/math] from linearity.

 

If you are assuming a more general linearity over a field, then as the rationals are only an algebra over themselves, they have no derivations. If you want algebras over rings, then the integers also work.

=Uncool-

Edited July 25, 2011 by uncool

[ajb]

So we define [math]\delta(1)= k[/math], then [math]\delta(x 1) = \delta(x)1 + x \delta(1) = \delta(x) + k x = \delta(x)[/math], so [math]k=0[/math]. So the identity is a "constant".

 

This is fine. Your a's and b's are real numbers, and these are "constants" [math]\delta(a)=0[/math] etc.

 

You can think of the reals as a real algebra over itself. But in this case thinking of it as a ring would be better. Which I think, because of my opening line above, requires [math]\delta(1) =0[/math].