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1. • CommentRowNumber1.
   • CommentAuthorJohn Baez
   • CommentTimeDec 8th 2010
   • PermaLink
   Author: John Baez
   Format: MarkdownItexI started a stub [[symmetric space]] and added a bit of overlapping material to [[quandle]]. I would like to talk about Lie and Jordan triple systems, but I need this introductory material first.

   I started a stub

   symmetric space

   and added a bit of overlapping material to quandle. I would like to talk about Lie and Jordan triple systems, but I need this introductory material first.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeDec 8th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexI added some links.

   I added some links.

3. • CommentRowNumber3.
   • CommentAuthorTobyBartels
   • CommentTimeDec 12th 2010
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexI put a disambiguation block at the top.

   I put a disambiguation block at the top.

4. • CommentRowNumber4.
   • CommentAuthorzskoda
   • CommentTimeSep 8th 2011
   • (edited Sep 8th 2011)
   • PermaLink
   Author: zskoda
   Format: MarkdownItex[[John Baez]] says in the entry that a [[symmetric space]] is a special case of a smooth [[quandle]]. I have Loos's book (Russian translation) for over 20 years and always considered it complicated so never went into details, but I have a slight problem with that elementary statement. According to standard literature, including Loos, a symmetric space is (equivalent to) a smooth manifold $M$ with multiplication $\cdot : M\times M\to M$ which is a smooth map such that for all $x,y,z\in M$ 1. $x \cdot x = x$ 2. $x \cdot (x\cdot y) = y$ 3. $x\cdot (y \cdot z) = (x \cdot y)\cdot (x \cdot z)$ 4. for every $x$ there is a neighborhood $U\subset M$ such that $x\cdot y = y$ implies $x = y$ for all $y\in U$. (Edit: I have now copied this approach verbatim to [[symmetric space]]). Being a [[quandle]] means that 1 and 3 are satisfied and * (1a) for all $x,y\in M$ there is $z\in M$ such that $x \cdot z = y$. See e.g. Kinyon's [slides](http://www.cirm.univ-mrs.fr/videos/2008/exposes/284/Kinyon.pdf). Is it obvious that (1a) follows, or there is some disagreement in definitions ? There is a Cartan's model of a symmetric space via certain type of a homogeneous space but the operation is not the most obvious one, but still it is likely that it implies the above statement, but I am not sure. By the way how to generalize the additional condition (1a) in an arbitrary monoidal category, or at least, category with products (the entry mentions "quandle object").

   John Baez says in the entry that a symmetric space is a special case of a smooth quandle. I have Loos’s book (Russian translation) for over 20 years and always considered it complicated so never went into details, but I have a slight problem with that elementary statement.

   According to standard literature, including Loos, a symmetric space is (equivalent to) a smooth manifold $M$ with multiplication $\cdot : M\times M\to M$ which is a smooth map such that for all $x,y,z\in M$

   1. $x \cdot x = x$
   2. $x \cdot (x\cdot y) = y$
   3. $x\cdot (y \cdot z) = (x \cdot y)\cdot (x \cdot z)$
   4. for every $x$ there is a neighborhood $U\subset M$ such that $x\cdot y = y$ implies $x = y$ for all $y\in U$.

   (Edit: I have now copied this approach verbatim to symmetric space). Being a quandle means that 1 and 3 are satisfied and

   • (1a) for all $x,y\in M$ there is $z\in M$ such that $x \cdot z = y$. See e.g. Kinyon’s slides.

   Is it obvious that (1a) follows, or there is some disagreement in definitions ? There is a Cartan’s model of a symmetric space via certain type of a homogeneous space but the operation is not the most obvious one, but still it is likely that it implies the above statement, but I am not sure.

   By the way how to generalize the additional condition (1a) in an arbitrary monoidal category, or at least, category with products (the entry mentions “quandle object”).

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 8th 2011
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexZoran, I haven't looked at your sources, but the way you've written it down, axiom 2 directly implies (1a). So I'm not quite sure what you're asking. Also, do you want unique existentiation in (1a)?

   Zoran, I haven’t looked at your sources, but the way you’ve written it down, axiom 2 directly implies (1a). So I’m not quite sure what you’re asking. Also, do you want unique existentiation in (1a)?

6. • CommentRowNumber6.
   • CommentAuthorzskoda
   • CommentTimeSep 8th 2011
   • (edited Sep 8th 2011)
   • PermaLink
   Author: zskoda
   Format: MarkdownItexOh I was stupid, thanks Todd! But still I am not clear about this. Yes, it should be the **unique** existence according to the entry [[rack]] (in fact one has not only right quasigroup but a [[quasigroup]]). Quandle is an idempotent rack, by the very definition. Rack may be restated as a left selfdistributive magma with right invertibility. Thus it should have been stated * (1a) for all $x,y\in M$ there is a unique $z\in M$ such that $x \cdot z = y$. But then the discreteness requirement 4, or John's phrase that "each point $a\in M$ is the isolated fixed point of $a\cdot - : M\to M$" looks superfluous in view of the uniqueness. John has a remark in the bibliography section > The definition in terms of quandles coincides with the classical definition in the case of connected symmetric spaces. which intuitively looks like coupled with discreteness gives uniqueness (probably with a little more work with geometry here). But anyway quandle means uniqueness and then John's discreteness as well as 4 in the Loos' definition are superfluous. I am still confused.

   Oh I was stupid, thanks Todd! But still I am not clear about this. Yes, it should be the unique existence according to the entry rack (in fact one has not only right quasigroup but a quasigroup). Quandle is an idempotent rack, by the very definition. Rack may be restated as a left selfdistributive magma with right invertibility. Thus it should have been stated

   • (1a) for all $x,y\in M$ there is a unique $z\in M$ such that $x \cdot z = y$.

   But then the discreteness requirement 4, or John’s phrase that “each point $a\in M$ is the isolated fixed point of $a\cdot - : M\to M$” looks superfluous in view of the uniqueness.

   John has a remark in the bibliography section

   The definition in terms of quandles coincides with the classical definition in the case of connected symmetric spaces.

   which intuitively looks like coupled with discreteness gives uniqueness (probably with a little more work with geometry here). But anyway quandle means uniqueness and then John’s discreteness as well as 4 in the Loos’ definition are superfluous. I am still confused.

7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 8th 2011
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexSorry, Zoran. I have two questions: first, is your condition 4 what you meant to write? I see $z$ coming out of nowhere, as it were. Should that have been a $y$? Second, I don't understand how the isolated fixed point condition becomes superfluous. Surely there can be multiple $b$ such that $a \cdot b = b$ (think of the rack given by a group), so I don't see how the condition becomes superfluous.

   Sorry, Zoran. I have two questions: first, is your condition 4 what you meant to write? I see $z$ coming out of nowhere, as it were. Should that have been a $y$? Second, I don’t understand how the isolated fixed point condition becomes superfluous. Surely there can be multiple $b$ such that $a \cdot b = b$ (think of the rack given by a group), so I don’t see how the condition becomes superfluous.

8. • CommentRowNumber8.
   • CommentAuthorzskoda
   • CommentTimeSep 8th 2011
   • (edited Sep 8th 2011)
   • PermaLink
   Author: zskoda
   Format: MarkdownItexIt is indeed $y$, I corrected now above. I am all the time changing (as I need both) between US and Croatian keyboard map and $y$ and $z$ permute in a role of a critical pair :) You are perfectly right, $b$ is also varying (what was I overlooking in thinking, so erratic today!) so it is OK. 4 is hence not superfluous! So now, I am just left with seeing the proof that > The definition in terms of quandles coincides with the classical definition in the case of __connected__ symmetric spaces. New entry [[self-distributivity]].

   It is indeed $y$, I corrected now above. I am all the time changing (as I need both) between US and Croatian keyboard map and $y$ and $z$ permute in a role of a critical pair :)

   You are perfectly right, $b$ is also varying (what was I overlooking in thinking, so erratic today!) so it is OK. 4 is hence not superfluous! So now, I am just left with seeing the proof that

   The definition in terms of quandles coincides with the classical definition in the case of connected symmetric spaces.

   New entry self-distributivity.

9. • CommentRowNumber9.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 8th 2011
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexBy the way, someone should put in at some point the connection between quandles and large cardinal axioms. (Richard Borcherds had a post on this on his blog.) I'm really not the best person to be discussing symmetric spaces with; the most I can do right now is rely on wikipedia and the nLab articles and the only book I have at home (Helgason's Differential Geometry, Lie Groups, and Symmetric Spaces). But let me see, just for the sake of the exercise, how far I can go with the proof you want. By a symmetric space according to Cartan, I guess you mean a homogeneous space $G/H$ where $G$ is a Lie group equipped with an involution $\sigma: G \to G$ and $H$ is an open subgroup of the group of fixed points of $\sigma$. I didn't see any definition of the quandle structure for this example on any of the nLab pages, but reading Wikipedia, I deduced that it must be $$(g H) \cdot (g'H) \coloneqq g \sigma(g^{-1}g') H$$ and it seems pretty straightforward to verify the equational axioms for an involutory quandle. So I take it that you want a proof of the isolated fixed point condition. I admit I don't see it offhand; did you already look over the references? (Loos or Wolfgang Bertram.) For the other direction, going from an involutory quandle object in manifolds (plus the isolated fixpoint condition) to the homogeneous space representation, I can only guess that the Lie group $G$ is supposed to be the smooth automorphism group of the quandle structure, and $H$ is an isotropy subgroup at a chosen point. Without thinking about it hard, I would guess the isolated fixpoint condition would aid in showing $H$ to be open, and I suppose the quandle structure will be put to use to construct the involution of $G$. I'm not sure I'm ready to go further right now, but is this on the right track, do you think?

   By the way, someone should put in at some point the connection between quandles and large cardinal axioms. (Richard Borcherds had a post on this on his blog.)

   I’m really not the best person to be discussing symmetric spaces with; the most I can do right now is rely on wikipedia and the nLab articles and the only book I have at home (Helgason’s Differential Geometry, Lie Groups, and Symmetric Spaces). But let me see, just for the sake of the exercise, how far I can go with the proof you want.

   By a symmetric space according to Cartan, I guess you mean a homogeneous space $G/H$ where $G$ is a Lie group equipped with an involution $\sigma: G \to G$ and $H$ is an open subgroup of the group of fixed points of $\sigma$. I didn’t see any definition of the quandle structure for this example on any of the nLab pages, but reading Wikipedia, I deduced that it must be

   $$(g H) \cdot (g'H) \coloneqq g \sigma(g^{-1}g') H$$

   and it seems pretty straightforward to verify the equational axioms for an involutory quandle. So I take it that you want a proof of the isolated fixed point condition. I admit I don’t see it offhand; did you already look over the references? (Loos or Wolfgang Bertram.)

   For the other direction, going from an involutory quandle object in manifolds (plus the isolated fixpoint condition) to the homogeneous space representation, I can only guess that the Lie group $G$ is supposed to be the smooth automorphism group of the quandle structure, and $H$ is an isotropy subgroup at a chosen point. Without thinking about it hard, I would guess the isolated fixpoint condition would aid in showing $H$ to be open, and I suppose the quandle structure will be put to use to construct the involution of $G$. I’m not sure I’m ready to go further right now, but is this on the right track, do you think?

10. • CommentRowNumber10.
    • CommentAuthorTobyBartels
    • CommentTimeSep 8th 2011
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItex>New entry [[self-distributive operation|self-distributivity]]. I added some examples and a link from [[rack]].

    New entry self-distributivity.

    I added some examples and a link from rack.

11. • CommentRowNumber11.
    • CommentAuthorzskoda
    • CommentTimeSep 9th 2011
    • PermaLink
    Author: zskoda
    Format: MarkdownItexThanks Toby! Yes, I am roughly somewhere there. Maybe I have too many references at hand, all very exhaustive, and it is a bit hard to isolate the basic pieces out. Probably I will sort it completely over the weekend.

    Thanks Toby! Yes, I am roughly somewhere there. Maybe I have too many references at hand, all very exhaustive, and it is a bit hard to isolate the basic pieces out. Probably I will sort it completely over the weekend.