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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeApr 22nd 2010
   • (edited Oct 30th 2013)
   • PermaLink
   Author: Urs
   Format: MarkdownItexChris Schommer-Pries posted a question/suggestion in the query box at [[semisimple category]]

   Chris Schommer-Pries posted a question/suggestion in the query box at semisimple category

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeOct 30th 2013
   • (edited Oct 30th 2013)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI have touched the formatting at _[[semisimple category]]_ slightly, and hereby I am moving that old query box from there to here: **** begin forwarded discussion **** What does it mean to 'have subobjects'? (I assume that the 'direct sums' are [[biproduct]]s.) ---Toby have subobjects = idempotents split and yes, finite biproducts. Simple objects are ones in which End(X) = k. [[Urs Schreiber|Urs]]: shouldn't we say something like: a category is semisimple if each object is a direct sum of finitely many simple objects? [[Bruce Bartlett|Bruce]]: Urs, you're right, and that's indeed the way one morally thinks about it, but it's a less canonical way of proceeding. We ask ourselves: given a linear category with direct sums and subobjects, and a chosen maximal collection $\{X_i\}$ of nonisomorphic simple objects, how can we check if its semisimple? In the one way, we have to check whether a certain canonically defined map is an isomorphism. In the other way, we have to check if each object $V$ can be expressed as a direct sum of the $X_i$'s. Actually finding such a decomposition would be a noncanonical operation. So your shorter more snappy definition would force an auditor to perform an evil thing if he actually wanted to check it :-) It's that old thing about "only that part of a representation which behaves like an irreducible $\rho$ is canonical, the actual break-down of that rep into direct sums of $\rho$'s is noncanonical". That is, what is canonical is $Hom(V, X_i)$ and not $V = \bigoplus_i n_i X_i$. Also, if we just had "a category is semisimple if each object is a direct sum of finitely many simple objects" without the conditions on direct sums and subobjects then we could have someone who nastily removes, say, all three-dimensional vector spaces from $Vect$. It would still satisfy "each object is a direct sum of finitely many simple objects" but it shouldnt be regarded as a semisimple category since there are "holes". [[Urs Schreiber|Urs]]: okay, so "each object is finite sum of simples" is the _right idea_, while the _right defintion_ is a bit different. I have accoridngly now created a section "Idea" with the former statement. (Every entry should start with a section "Idea"!) See if you like this. Otherwise, feel free to adjust. But in any case, it would be nice to have a discussion on how the "right definition" _implies_ that every object is isomorphic to a finite direct sum of representables. __[[Chris Schommer-Pries]]__: Shouldn't you require that for simples, End(X,X) is a simple algebra, not necessarily the ground field? For example the category of H-modules where H is the quaternion algebra over the reals. Shouldn't this be semi-simple? [[Christopher Douglas|Chris]]: It is not standard to assume the direct sum decomposition is finite in defining a semisimple category. For instance, the category of all vector spaces is semisimple. What's written here at present looks like a definition of semisimple for categories where it has already been assumed that every object has finite length. **** end forwarded discussion ****

   I have touched the formatting at semisimple category slightly, and hereby I am moving that old query box from there to here:

   ---

   begin forwarded discussion

   ---

   What does it mean to ’have subobjects’? (I assume that the ’direct sums’ are biproducts.) —Toby

   have subobjects = idempotents split and yes, finite biproducts. Simple objects are ones in which End(X) = k.

   Urs: shouldn’t we say something like: a category is semisimple if each object is a direct sum of finitely many simple objects?

   Bruce: Urs, you’re right, and that’s indeed the way one morally thinks about it, but it’s a less canonical way of proceeding. We ask ourselves: given a linear category with direct sums and subobjects, and a chosen maximal collection $\{X_i\}$ of nonisomorphic simple objects, how can we check if its semisimple? In the one way, we have to check whether a certain canonically defined map is an isomorphism. In the other way, we have to check if each object $V$ can be expressed as a direct sum of the $X_i$’s. Actually finding such a decomposition would be a noncanonical operation. So your shorter more snappy definition would force an auditor to perform an evil thing if he actually wanted to check it :-) It’s that old thing about “only that part of a representation which behaves like an irreducible $\rho$ is canonical, the actual break-down of that rep into direct sums of $\rho$’s is noncanonical”. That is, what is canonical is $Hom(V, X_i)$ and not $V = \bigoplus_i n_i X_i$.

   Also, if we just had “a category is semisimple if each object is a direct sum of finitely many simple objects” without the conditions on direct sums and subobjects then we could have someone who nastily removes, say, all three-dimensional vector spaces from $Vect$. It would still satisfy “each object is a direct sum of finitely many simple objects” but it shouldnt be regarded as a semisimple category since there are “holes”.

   Urs: okay, so “each object is finite sum of simples” is the right idea, while the right defintion is a bit different. I have accoridngly now created a section “Idea” with the former statement. (Every entry should start with a section “Idea”!) See if you like this. Otherwise, feel free to adjust.

   But in any case, it would be nice to have a discussion on how the “right definition” implies that every object is isomorphic to a finite direct sum of representables.

   Chris Schommer-Pries: Shouldn’t you require that for simples, End(X,X) is a simple algebra, not necessarily the ground field? For example the category of H-modules where H is the quaternion algebra over the reals. Shouldn’t this be semi-simple?

   Chris: It is not standard to assume the direct sum decomposition is finite in defining a semisimple category. For instance, the category of all vector spaces is semisimple. What’s written here at present looks like a definition of semisimple for categories where it has already been assumed that every object has finite length.

   ---

   end forwarded discussion

   ---

3. • CommentRowNumber3.
   • CommentAuthorTobyBartels
   • CommentTimeOct 31st 2013
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexI agree with Chris at the end here. We should not restrict ourselves to finite sums, although that might be an interesting variation. ($\kappa$-semisimple, anybody?)

   I agree with Chris at the end here. We should not restrict ourselves to finite sums, although that might be an interesting variation. ($\kappa$-semisimple, anybody?)

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeOct 31st 2013
   • PermaLink
   Author: Urs
   Format: MarkdownItexThere are different traditions. In references like those listed at [[FFRS-formalism]] "semisimple" implies the finiteness clause. (Though just checking with their latest I see that they did switch to saying "finitely semismple" now.)

   There are different traditions. In references like those listed at FFRS-formalism “semisimple” implies the finiteness clause.

   (Though just checking with their latest I see that they did switch to saying “finitely semismple” now.)

5. • CommentRowNumber5.
   • CommentAuthorTobyBartels
   • CommentTimeNov 1st 2013
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItex>they did switch to saying "finitely semismple" That seems like a reasonable use of language. Of course, in some contexts, that could be the default, just like ‘ring’ means something commutative to some people.

   they did switch to saying “finitely semismple”

   That seems like a reasonable use of language. Of course, in some contexts, that could be the default, just like ‘ring’ means something commutative to some people.