Timeline for Moore-Penrose Inverse as an adjoint
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2012 at 17:25 | vote | accept | Nicolas Schmidt | ||
| Jun 9, 2012 at 14:17 | answer | added | Michal R. Przybylek | timeline score: 6 | |
| Jun 9, 2012 at 11:20 | comment | added | Michal R. Przybylek | Here is the final answer --- in any 2-category adjunctions composes, i.e. if $f \dashv g$ and $h \dashv k$ for compatible $f$ and $h$ then $f h \dashv k g$, whreas Moore-Penrose pseudoinverses generally do not. | |
| Jun 8, 2012 at 13:03 | comment | added | Michal R. Przybylek | The page also says "any adjunction between posets is idempotent", but this refers to posets thought as of categories, which has nothing to do with our 2-posets. | |
| Jun 8, 2012 at 13:00 | comment | added | Michal R. Przybylek | Nicolas, I do not buy your explanation. I do not know much about "indempotent adjunction", but have found the following page: ncatlab.org/nlab/show/idempotent+adjunction. It seems that the idea behind the indempotent adjunction is to impose morphisms $f \rightarrow fgf$ and $g \rightarrow gfg$ to be isomorphisms. If you assume furthermore that these isomorphisms are identities, then you get your equations --- not "for free", but directly from your definition. | |
| Jun 7, 2012 at 19:50 | comment | added | Nicolas Schmidt | The equality $f = fgf$ is related to the triangle equations through the notion of "idempotent adjunction", a particularly nice class of adjunctions. Whenever you have an idempotent adjunction, you get the familiar equations $f = fgf$ and $g = gfg$ (interpreted suitably) for free. For poset-categories any adjunction is automatically idempotent, hence my suggestion, but even outside this context many adjunctions "in nature" turn out to be idempotent. | |
| Jun 5, 2012 at 18:36 | comment | added | Michal R. Przybylek | b) suppose that $\mathbb{W}$ is a 2-category buit upon $\mathbf{Vect}$ where every pseudoinverse is a part of an adjuntion (satisfying, perhaps, some other conditions), then every morphism in $\mathbb{W}$ has both left and right adjoint, furthermore these adjoint functors are isomorphic to the pseudoinverse (so isomorphic to each other); such $\mathbb{W}$ would be really a strange 2-category. | |
| Jun 5, 2012 at 18:35 | comment | added | Michal R. Przybylek | Nicolas, I am really skeptical about any such construction. Let me summarize two reasons: a) the triangle equalities say that the composition of the obvious morphisms $f \to fgf \to f$ and $g \to gfg \to g$ are identities; this has about nothing to do with your equations; it turns out that if we restrict our 2-categories to 2-posets, then your equations are satisfied, but not because of the triangle equalities, but because of a mere existence of 2-morphisms $\mathit{id} \to gf$ and $fg \to \mathit{id}$ | |
| Jun 5, 2012 at 15:29 | comment | added | Nicolas Schmidt | I guess the construction I suggested really is a dead-end. Maybe the answer to the weaker version of my question still is "yes"? Somehow I can't get this tempting idea out of my head. | |
| Jun 5, 2012 at 9:41 | comment | added | Michal R. Przybylek | Now let us assume that we are in a 2-poset and $f$ is both left and right adjoint to $g$. From the first adjunction we have inequalities $\mathit{id} \le gf$, $fg \le \mathit{id}$, and from the second $\mathit{id} \le fg$, $gf \le \mathit{id}$. Thus $fg = \mathit{id}$ and $gf = \mathit{id}$, so the concept of adjoint sequence $f \dashv g \dashv f$ collapses to the concept of inverse. | |
| Jun 5, 2012 at 9:41 | comment | added | Michal R. Przybylek | Chris, it will make things even worse --- adjoint morphisms are unique up to 2-isomorphisms, and only 2-isomorphisms in a 2-poset are identities. The real problem is in the symmetry of the definition. The definition of a Moore-Penrose pseudoinverse is completly symmetric for $f$ and $g$. This means that if $f$ had been left (right) adjoint to $g$, then it would have automatically been also right (resp. left) adjoint to $g$. | |
| Jun 5, 2012 at 8:50 | comment | added | Chris Heunen | If you drop the latter two conditions, pseudo-inverses need not be unique anymore. That might get around the obstruction Michal raises. | |
| Jun 4, 2012 at 20:24 | comment | added | Michal R. Przybylek | If I understand your question, the answer is: no. The pseudoinverse of the pseudoinverse of a map is the same map, so the adjunction $f \dashv g$ implies the adjoint sequence $f \dashv g \dashv f$. Thus we have $fg \le \mathit{id}$ and $fg \ge \mathit{id}$, and by antisymmetry of partial orders, $fg=\mathit{id}$. | |
| Jun 4, 2012 at 15:58 | history | asked | Nicolas Schmidt | CC BY-SA 3.0 |