Dgdrxrt

Why are there no referendums in the US?

What happens if you roll doubles 3 times then land on "Go to jail?"

India just shot down a satellite from the ground. At what altitude range is the resulting debris field?

Is expanding the research of a group into machine learning as a PhD student risky?

How does it work when somebody invests in my business?

Where does the Z80 processor start executing from?

What can we do to stop prior company from asking us questions?

How does the UK government determine the size of a mandate?

Do the temporary hit points from Reckless Abandon stack if I make multiple attacks on my turn?

How to pronounce the slash sign

Large drywall patch supports

What is the difference between "behavior" and "behaviour"?

Avoiding estate tax by giving multiple gifts

Risk of infection at the gym?

Closest Prime Number

Go Pregnant or Go Home

Pole-zeros of a real-valued causal FIR system

How do I go from 300 unfinished/half written blog posts, to published posts?

What Brexit proposals are on the table in the indicative votes on the 27th of March 2019?

Proof of work - lottery approach

Class Action - which options I have?

Implement the Thanos sorting algorithm

How to write papers efficiently when English isn't my first language?

How to Reset Passwords on Multiple Websites Easily?

k-linear categories

Kan extensions for linear categoriesWhat are presentable categories?Distributivity in linear monoidal categoriesYoneda lemma for enriched categoriesIsomorphic categoriesAre functors (from small categories) functions?Is the enriched $underlinemathrmTop_mathrmG$ tensored and cotensored?k-linear categoryFor which algebra $A$: $mathrmvect cong A-$modIsomorphism between Hom-space and $k$

0

$begingroup$

I have read several times about $k$-linear categories. I understand in general the construction of enriched categories.

But in the special case of vector spaces I am confused about the vector space addition.
If the Hom spaces are vector spaces, where does the addition come from? Composing isn't commutative, so that doesn't work...

In particular, I have a semisimple category $mathcalC$ which is a module category over $mathrmvect_k$ - meaning there exist an action on both objects $V.C$ ($V in mathrmvect$ and $C in mathcalC$) and morphisms $f.phi$ with $f$ vector morphism and $phi in mathcalC$.

I want to show that this is $k$-linear i.e. that $mathrmHom(C, C')$ is a vector space $forall C, C' in mathcalC$.

I got the scalar multiplication with help of the action on morphisms, but how does one get the addition?

Thanks in advance

vector-spaces category-theory

share|cite|improve this question

asked yesterday

P. SchulzeP. Schulze

719

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Isn't your semisimple category already additive?
  $endgroup$
  – Kevin Carlson
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

I have read several times about $k$-linear categories. I understand in general the construction of enriched categories.

But in the special case of vector spaces I am confused about the vector space addition.
If the Hom spaces are vector spaces, where does the addition come from? Composing isn't commutative, so that doesn't work...

In particular, I have a semisimple category $mathcalC$ which is a module category over $mathrmvect_k$ - meaning there exist an action on both objects $V.C$ ($V in mathrmvect$ and $C in mathcalC$) and morphisms $f.phi$ with $f$ vector morphism and $phi in mathcalC$.

I want to show that this is $k$-linear i.e. that $mathrmHom(C, C')$ is a vector space $forall C, C' in mathcalC$.

I got the scalar multiplication with help of the action on morphisms, but how does one get the addition?

Thanks in advance

vector-spaces category-theory

share|cite|improve this question

asked yesterday

P. SchulzeP. Schulze

719

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Isn't your semisimple category already additive?
  $endgroup$
  – Kevin Carlson
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

I have read several times about $k$-linear categories. I understand in general the construction of enriched categories.

But in the special case of vector spaces I am confused about the vector space addition.
If the Hom spaces are vector spaces, where does the addition come from? Composing isn't commutative, so that doesn't work...

In particular, I have a semisimple category $mathcalC$ which is a module category over $mathrmvect_k$ - meaning there exist an action on both objects $V.C$ ($V in mathrmvect$ and $C in mathcalC$) and morphisms $f.phi$ with $f$ vector morphism and $phi in mathcalC$.

I want to show that this is $k$-linear i.e. that $mathrmHom(C, C')$ is a vector space $forall C, C' in mathcalC$.

I got the scalar multiplication with help of the action on morphisms, but how does one get the addition?

Thanks in advance

vector-spaces category-theory

share|cite|improve this question

asked yesterday

P. SchulzeP. Schulze

719

$endgroup$

share|cite|improve this question

asked yesterday

P. SchulzeP. Schulze

719

share|cite|improve this question

asked yesterday

P. SchulzeP. Schulze

719

asked yesterday

P. SchulzeP. Schulze

719

asked yesterday

P. SchulzeP. Schulze

719

1 Answer
1

active

oldest

votes

1

$begingroup$

In general, enrichment is extra structure, and you need to specify it. It's worth understanding in detail the one-object case: for $V$ a monoidal category (such as vector spaces with tensor product), a one-object $V$-enriched category is a monoid in $V$. So, for example, a one-object $k$-linear category is a $k$-algebra. It has an underlying category which is given by forgetting the vector space structure and remembering only the multiplication; you need to specify the vector space structure, and in particular the addition, as extra data on top of this.

However, enrichment over commutative monoids has a very peculiar property, which is that under mild hypotheses (the existence of either finite products or finite coproducts) it is actually determined by the bare category structure. See this blog post for a lengthy discussion of how this works.

share|cite|improve this answer

answered 19 hours ago

Qiaochu YuanQiaochu Yuan

281k32595940

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3163331%2fk-linear-categories%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

In general, enrichment is extra structure, and you need to specify it. It's worth understanding in detail the one-object case: for $V$ a monoidal category (such as vector spaces with tensor product), a one-object $V$-enriched category is a monoid in $V$. So, for example, a one-object $k$-linear category is a $k$-algebra. It has an underlying category which is given by forgetting the vector space structure and remembering only the multiplication; you need to specify the vector space structure, and in particular the addition, as extra data on top of this.

However, enrichment over commutative monoids has a very peculiar property, which is that under mild hypotheses (the existence of either finite products or finite coproducts) it is actually determined by the bare category structure. See this blog post for a lengthy discussion of how this works.

share|cite|improve this answer

answered 19 hours ago

Qiaochu YuanQiaochu Yuan

281k32595940

$endgroup$

add a comment | 

1

$begingroup$

In general, enrichment is extra structure, and you need to specify it. It's worth understanding in detail the one-object case: for $V$ a monoidal category (such as vector spaces with tensor product), a one-object $V$-enriched category is a monoid in $V$. So, for example, a one-object $k$-linear category is a $k$-algebra. It has an underlying category which is given by forgetting the vector space structure and remembering only the multiplication; you need to specify the vector space structure, and in particular the addition, as extra data on top of this.

However, enrichment over commutative monoids has a very peculiar property, which is that under mild hypotheses (the existence of either finite products or finite coproducts) it is actually determined by the bare category structure. See this blog post for a lengthy discussion of how this works.

share|cite|improve this answer

answered 19 hours ago

Qiaochu YuanQiaochu Yuan

281k32595940

$endgroup$

add a comment | 

$begingroup$

In general, enrichment is extra structure, and you need to specify it. It's worth understanding in detail the one-object case: for $V$ a monoidal category (such as vector spaces with tensor product), a one-object $V$-enriched category is a monoid in $V$. So, for example, a one-object $k$-linear category is a $k$-algebra. It has an underlying category which is given by forgetting the vector space structure and remembering only the multiplication; you need to specify the vector space structure, and in particular the addition, as extra data on top of this.

However, enrichment over commutative monoids has a very peculiar property, which is that under mild hypotheses (the existence of either finite products or finite coproducts) it is actually determined by the bare category structure. See this blog post for a lengthy discussion of how this works.

share|cite|improve this answer

answered 19 hours ago

Qiaochu YuanQiaochu Yuan

281k32595940

$endgroup$

share|cite|improve this answer

answered 19 hours ago

Qiaochu YuanQiaochu Yuan

281k32595940

answered 19 hours ago

Qiaochu YuanQiaochu Yuan

281k32595940

answered 19 hours ago

Qiaochu YuanQiaochu Yuan

281k32595940