From Apstheory
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- | ==Definition==
| + | AC5gQV I cannot thank you enough for the blog article.Really looking forward to read more. Fantastic. |
- | Let <math>(G,\\Phi)</math> be an [[APS]] over a monoidal concrete category. Then a sub-APS of <math>(G,\\Phi)</math> associates to each <math>n</math> a subobject <math>H_n</math> of <math>G_n</math> such that the restriction of <math>\\Phi_{m,n}</math> to <math>H_m</math> × <math>H_n</math> takes it inside <math>H_{m+n}</math>.
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- | Thus we can view <math>(H,\\Phi)</math> as an APS in its own right (note that since the associativity condition is satisfied for the block concatenation on <math>G</math>, it is also satisfied for the block concatenation on <math>H</math>.
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- | Since the <math>\\Phi</math> is understood for the sub-APS, we may omit it and simply say that <math>H</math> is a sub-APS of <math>G</math>.
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- | ==Other notions==
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- | ===Smallest sub-APS containing a given collection of subobjects===
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- | If we are given a subobject of <math>G_n</math> for every <math>n</math>, we can talk of the sub-APS generated by these subobjects. This associates to every <math>n</math>, the set of elements in <math>G_n</math> that arise via a block concatenation of elements upto <math>n</math>.
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- | ===Intersection of sub-APSes===
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- | Given two sub-APSes <math>H</math> and <math>K</math> of an APS <math>G</math>, the intersection of <math>H</math> and <math>K</math> associates to each <math>n</math> the subset <math>L_n = H_n</math> ∩ <math>K_n</math> of <math>G_n</math>. If the intersection of two subobjects is a subobject (which, for instance, is the case in algebraic structures) then <math>L</math> is also a sub-APS of <math>G</math>.
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- | Similarly, if an arbitrary intersection of subobjects is a subobject, then an arbitrary intersection of sub-APSes is a sub-APS.
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Current revision as of 18:55, 14 December 2013
This article gives a basic definition in the APS theory. It is strictly local to the wiki
AC5gQV I cannot thank you enough for the blog article.Really looking forward to read more. Fantastic.