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1. • CommentRowNumber1.
   • CommentAuthorJohn Baez
   • CommentTimeAug 10th 2016
   • (edited Aug 10th 2016)
   • PermaLink
   Author: John Baez
   Format: MarkdownItexI made some progress understanding the concept of [[Albanese variety]] and updated this page. The miracle here is that being an abelian variety is just a _property_ of a pointed connected projective algebraic variety, not extra structure. As Qiaochu Yuan pointed out on MathOverflow, any basepoint-preserving continuous map between tori is _homotopic_ to a group homomorphism. But when these tori are abelian varieties, and the map preserves that structure, it's actually _equal_ to a group homomorphism!

   I made some progress understanding the concept of Albanese variety and updated this page.

   The miracle here is that being an abelian variety is just a property of a pointed connected projective algebraic variety, not extra structure.

   As Qiaochu Yuan pointed out on MathOverflow, any basepoint-preserving continuous map between tori is homotopic to a group homomorphism. But when these tori are abelian varieties, and the map preserves that structure, it’s actually equal to a group homomorphism!

2. • CommentRowNumber2.
   • CommentAuthorfastlane69
   • CommentTimeAug 10th 2016
   • (edited Aug 10th 2016)
   • PermaLink
   Author: fastlane69
   Format: MarkdownItexSorry for what may be naive questions or trivial restatement of the results: Does this mean that if I have a torus and I deform it into another torus, as long as I preserve the basepoint during this transformation, i.e., hold onto the basepoint while I muck about the rest of the tori (is that a good way to think of "basepoint preserving map"?) then the resultant maps between tori are guaranteed to "look and behave" (is this a good way of thinking of homotopic?) like group homs but if the original tori is abelian, the resultant maps "are" group homs and the resultant tori is also abelian? And further, by you stating this as a property and not a extra structure, that ANY model we can put into the framework of a pointed connected projective algebraic variety (is this the same as a pointed algebraic manifold?) will automatically be abelian as will any basepoint-preserving transformations within it?

   Sorry for what may be naive questions or trivial restatement of the results:

   Does this mean that if I have a torus and I deform it into another torus, as long as I preserve the basepoint during this transformation, i.e., hold onto the basepoint while I muck about the rest of the tori (is that a good way to think of “basepoint preserving map”?) then the resultant maps between tori are guaranteed to “look and behave” (is this a good way of thinking of homotopic?) like group homs but if the original tori is abelian, the resultant maps “are” group homs and the resultant tori is also abelian?

   And further, by you stating this as a property and not a extra structure, that ANY model we can put into the framework of a pointed connected projective algebraic variety (is this the same as a pointed algebraic manifold?) will automatically be abelian as will any basepoint-preserving transformations within it?

3. • CommentRowNumber3.
   • CommentAuthorfastlane69
   • CommentTimeAug 10th 2016
   • (edited Aug 10th 2016)
   • PermaLink
   Author: fastlane69
   Format: MarkdownItexWhy I ask: In studying how a chromatic scale breaks down to the major scale, I view it as the symmetry breaking from an abelian cyclic Z12 group to a periodic abelian 2-generator Z7 group. Problem is that since Z7 is not a proper subgroup of Z12, there is no guarantee that because the parent Z12 scale is abelian that the child Z7 scale will be as well. However, if we view both Z12 and Z7 as tori then, if I understand this correctly, I can say that as long as I preserve the basepoint (i.e., fix my root on the note C; the $e^0$ element of Z12), then I can create any child tori I want and as long as I preserve the basepoint(i.e., keep my root at C for any child tori), then any child tori, Z7 included, is guaranteed to be abelian. IF I understand this correctly, this would be the missing link I've been searching for the last year or so in trying to find a connection between abelian cyclic Z12 and periodic abelian 2-generator Z7.

   Why I ask:

   In studying how a chromatic scale breaks down to the major scale, I view it as the symmetry breaking from an abelian cyclic Z12 group to a periodic abelian 2-generator Z7 group.

   Problem is that since Z7 is not a proper subgroup of Z12, there is no guarantee that because the parent Z12 scale is abelian that the child Z7 scale will be as well.

   However, if we view both Z12 and Z7 as tori then, if I understand this correctly, I can say that as long as I preserve the basepoint (i.e., fix my root on the note C; the $S^1/\mathbb{Z[e^0$ element of Z12), then I can create any child tori I want and as long as I preserve the basepoint(i.e., keep my root at C for any child tori), then any child tori, Z7 included, is guaranteed to be abelian.

   IF I understand this correctly, this would be the missing link I’ve been searching for the last year or so in trying to find a connection between abelian cyclic Z12 and periodic abelian 2-generator Z7.

4. • CommentRowNumber4.
   • CommentAuthorJohn Baez
   • CommentTimeAug 18th 2016
   • PermaLink
   Author: John Baez
   Format: MarkdownItexZ12 and Z7 are not tori, and they are not abelian varieties.

   Z12 and Z7 are not tori, and they are not abelian varieties.

5. • CommentRowNumber5.
   • CommentAuthorJohn Baez
   • CommentTimeAug 18th 2016
   • (edited Aug 19th 2016)
   • PermaLink
   Author: John Baez
   Format: MarkdownItexI added to [[abelian variety]] an explanation of how from Mumford's "rigidity lemma" we can derive the fact that 1) every group object in the category of connected projective varieties is an _abelian_ variety, and 2) every basepoint-preserving map between abelian varieties is a homomorphism. I also updated [[Albanese variety]] to reflect these changes. This is stuff some of us worked out at the nCafe thread [Two miracles in algebraic geometry](https://golem.ph.utexas.edu/category/2016/08/the_magic_of_algebraic_geometr.html). It turns out that there's one fundamental "miracle" here: the rigidity lemma. Just trying to learn a bit of algebraic geometry...

   I added to abelian variety an explanation of how from Mumford’s “rigidity lemma” we can derive the fact that 1) every group object in the category of connected projective varieties is an abelian variety, and 2) every basepoint-preserving map between abelian varieties is a homomorphism. I also updated Albanese variety to reflect these changes.

   This is stuff some of us worked out at the nCafe thread Two miracles in algebraic geometry. It turns out that there’s one fundamental “miracle” here: the rigidity lemma.

   Just trying to learn a bit of algebraic geometry…

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeAug 18th 2016
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexThanks!

   Thanks!

7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 18th 2016
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexI'm confused. This is over any field $k$? What if I take $V = S^3$ as defined in $\mathbb{P}^4(\mathbb{R})$ by the homogeneous equation $x_0^2 + x_1^2 + x_2^2 + x_3^2 = x_4^2$? Under the ordinary reading of "connected", this is a connected real projective algebraic group under quaternionic multiplication, which is definable over $\mathbb{R}$ using polynomials. What fine print am I overlooking?

   I’m confused. This is over any field $k$? What if I take $V = S^3$ as defined in $\mathbb{P}^4(\mathbb{R})$ by the homogeneous equation $x_0^2 + x_1^2 + x_2^2 + x_3^2 = x_4^2$? Under the ordinary reading of “connected”, this is a connected real projective algebraic group under quaternionic multiplication, which is definable over $\mathbb{R}$ using polynomials.

   What fine print am I overlooking?

8. • CommentRowNumber8.
   • CommentAuthorDavidRoberts
   • CommentTimeAug 18th 2016
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexWhat's the multiplication map? I can't see it.

   What’s the multiplication map? I can’t see it.

9. • CommentRowNumber9.
   • CommentAuthorfastlane69
   • CommentTimeAug 18th 2016
   • (edited Aug 18th 2016)
   • PermaLink
   Author: fastlane69
   Format: MarkdownItexThanks for the reply, John! >Z12 and Z7 are not tori, Help me clear up my confusion please; from [Wolframs Maths definition of a Torus](http://mathworld.wolfram.com/Torus.html) >A second definition for n-tori relates to dimensionality. In one dimension, a line bends into circle, giving the 1-torus. In two dimensions, a rectangle wraps to a usual torus, also called the 2-torus. In three dimensions, the cube wraps to form a 3-manifold, or 3-torus So Z12 and Z7, as I read it, are 1-Tori. Regardless, you say they are not Albenese and thus don't work for my purposes, but I'm a bit confused by your saying the cyclic Zn is not a tori. VERY likely an important lack of understanding on my part which is why I'm pursuing this and asking. All the same, is Z12 $\cong$ Z4xZ3 of an abelian variety? This is a true torus, a 2-torus, $T^2$, as I understand it.

   Thanks for the reply, John!

   Z12 and Z7 are not tori,

   Help me clear up my confusion please; from Wolframs Maths definition of a Torus

   A second definition for n-tori relates to dimensionality. In one dimension, a line bends into circle, giving the 1-torus. In two dimensions, a rectangle wraps to a usual torus, also called the 2-torus. In three dimensions, the cube wraps to form a 3-manifold, or 3-torus

   So Z12 and Z7, as I read it, are 1-Tori.

   Regardless, you say they are not Albenese and thus don’t work for my purposes, but I’m a bit confused by your saying the cyclic Zn is not a tori. VERY likely an important lack of understanding on my part which is why I’m pursuing this and asking.

   All the same, is Z12 $\cong$ Z4xZ3 of an abelian variety? This is a true torus, a 2-torus, $T^2$, as I understand it.

10. • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 18th 2016
    • (edited Aug 18th 2016)
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexSorry, David R, were you asking me re #7? I know you know how quaternionic multiplication works, but I'll write it out; maybe it will help someone point out what I'm missing. In affine coordinates, it's $$\langle (x_0, x_1, x_2, x_3), (y_0, y_1, y_2, y_3) \rangle \mapsto (x_0 y_0 - x_1 y_1 - x_2 y_2 - x_3 y_3, x_0 y_1 + x_1 y_0 + x_2 y_3 - x_3 y_2, x_2 y_0 + x_0 y_2 - x_1 y_3 + x_3 y_1, x_0 y_3 + x_3 y_0 + x_1 y_2 - x_2 y_1)$$ This can easily be written in projective coordinates by the usual homogeneizing trick (replace $x_i$ in the above by $x_i/x_4$ and $y_i$ by $y_i/y_4$, and then multiply the outputs by $x_4 y_4$ to clear denominators). Edit: why don't I just write it out, then? If I'm not mistaken, in projective coordinates it's $$\langle (x_0, x_1, x_2, x_3, x_4), (y_0, y_1, y_2, y_3, y_4) \rangle \mapsto (x_0 y_0 - x_1 y_1 - x_2 y_2 - x_3 y_3, x_0 y_1 + x_1 y_0 + x_2 y_3 - x_3 y_2, x_2 y_0 + x_0 y_2 - x_1 y_3 + x_3 y_1, x_0 y_3 + x_3 y_0 + x_1 y_2 - x_2 y_1, x_4 y_4)$$

    Sorry, David R, were you asking me re #7?

    I know you know how quaternionic multiplication works, but I’ll write it out; maybe it will help someone point out what I’m missing. In affine coordinates, it’s

    $$\langle (x_0, x_1, x_2, x_3), (y_0, y_1, y_2, y_3) \rangle \mapsto (x_0 y_0 - x_1 y_1 - x_2 y_2 - x_3 y_3, x_0 y_1 + x_1 y_0 + x_2 y_3 - x_3 y_2, x_2 y_0 + x_0 y_2 - x_1 y_3 + x_3 y_1, x_0 y_3 + x_3 y_0 + x_1 y_2 - x_2 y_1)$$

    This can easily be written in projective coordinates by the usual homogeneizing trick (replace $x_i$ in the above by $x_i/x_4$ and $y_i$ by $y_i/y_4$, and then multiply the outputs by $x_4 y_4$ to clear denominators).

    Edit: why don’t I just write it out, then? If I’m not mistaken, in projective coordinates it’s

    $$\langle (x_0, x_1, x_2, x_3, x_4), (y_0, y_1, y_2, y_3, y_4) \rangle \mapsto (x_0 y_0 - x_1 y_1 - x_2 y_2 - x_3 y_3, x_0 y_1 + x_1 y_0 + x_2 y_3 - x_3 y_2, x_2 y_0 + x_0 y_2 - x_1 y_3 + x_3 y_1, x_0 y_3 + x_3 y_0 + x_1 y_2 - x_2 y_1, x_4 y_4)$$
11. • CommentRowNumber11.
    • CommentAuthorDavidRoberts
    • CommentTimeAug 18th 2016
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexTodd, yes. It doesn't _feel_ like it should be a multiplication on the variety you describe, but I guess I'd just need to sit down and do the calculation.

    Todd,

    yes. It doesn’t feel like it should be a multiplication on the variety you describe, but I guess I’d just need to sit down and do the calculation.

12. • CommentRowNumber12.
    • CommentAuthorDavidRoberts
    • CommentTimeAug 18th 2016
    • (edited Aug 18th 2016)
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItex@fastlane Your sets $\mathbb{Z}/12$ and $\mathbb{Z}/7$ are finite sets, not tori in the sense of the article you link. You seem to be implicitly embedding these into the circle and then thinking of the circle (which is a 1-dimensional torus). They are very different things! As far as algebraic geometry goes, an abelian variety is a curve/surface/etc described by equations, and as far as I can tell are _connected_ (EDIT: yes, see [Wikipedia](https://en.wikipedia.org/wiki/Abelian_variety#Algebraic_definition)), in the sense that it doesn't fall into multiple pieces -- which your examples definitely do.

    @fastlane

    Your sets $\mathbb{Z}/12$ and $\mathbb{Z}/7$ are finite sets, not tori in the sense of the article you link. You seem to be implicitly embedding these into the circle and then thinking of the circle (which is a 1-dimensional torus). They are very different things!

    As far as algebraic geometry goes, an abelian variety is a curve/surface/etc described by equations, and as far as I can tell are connected (EDIT: yes, see Wikipedia), in the sense that it doesn’t fall into multiple pieces – which your examples definitely do.

13. • CommentRowNumber13.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 18th 2016
    • (edited Aug 18th 2016)
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexDavid, all I'm doing is writing out the definition of quaternionic multiplication. It's clear this restricts to a multiplication on the unit sphere by the following argument. Since quaternionic conjugation $v \mapsto \widebar{v}$ satisfies $\widebar{u \cdot v} = \widebar{v} \cdot \widebar{u}$, and the unit sphere $V = S^3$ is defined by $v \cdot \widebar{v} = 1$, we have for $u, v \in S^3$ $$u v \cdot \widebar{u v} = u (v \cdot \widebar{v}) \widebar{u} = u \cdot \widebar{u} = 1$$ so $u v$ lands in $S^3$. I suspect that there's a missing hypothesis somewhere on the page. Perhaps we need $k$ algebraically closed or something to get the rigidity lemma, or maybe "connected" in algebraic geometry doesn't mean what it means topologically, even in the case $k = \mathbb{R}$.

    David, all I’m doing is writing out the definition of quaternionic multiplication. It’s clear this restricts to a multiplication on the unit sphere by the following argument. Since quaternionic conjugation $v \mapsto \widebar{v}$ satisfies $\widebar{u \cdot v} = \widebar{v} \cdot \widebar{u}$, and the unit sphere $V = S^3$ is defined by $v \cdot \widebar{v} = 1$, we have for $u, v \in S^3$

    $$u v \cdot \widebar{u v} = u (v \cdot \widebar{v}) \widebar{u} = u \cdot \widebar{u} = 1$$

    so $u v$ lands in $S^3$.

    I suspect that there’s a missing hypothesis somewhere on the page. Perhaps we need $k$ algebraically closed or something to get the rigidity lemma, or maybe “connected” in algebraic geometry doesn’t mean what it means topologically, even in the case $k = \mathbb{R}$.

14. • CommentRowNumber14.
    • CommentAuthorDavidRoberts
    • CommentTimeAug 18th 2016
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexTodd, ok, sure. I guess I'm missing something too, since I can't see where the argument breaks down either.

    Todd, ok, sure. I guess I’m missing something too, since I can’t see where the argument breaks down either.

15. • CommentRowNumber15.
    • CommentAuthorDavidRoberts
    • CommentTimeAug 18th 2016
    • (edited Aug 18th 2016)
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexSurely base changing to the complex numbers would also give an abelian variety (since this should just be a pullback of schemes to Spec(C)), but then I'm doubtful that the result is connected. This doesn't disprove anything, but it is highly suspicious.

    Surely base changing to the complex numbers would also give an abelian variety (since this should just be a pullback of schemes to Spec(C)), but then I’m doubtful that the result is connected. This doesn’t disprove anything, but it is highly suspicious.

16. • CommentRowNumber16.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 18th 2016
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexYeah, I was thinking along these lines as well.

    Yeah, I was thinking along these lines as well.

17. • CommentRowNumber17.
    • CommentAuthorJohn Baez
    • CommentTimeAug 19th 2016
    • (edited Aug 19th 2016)
    • PermaLink
    Author: John Baez
    Format: MarkdownItexInteresting issue, Todd. I don't see any problem with your counterexample; perhaps an even easier one is $\mathbb{R}\mathrm{P}^3 \cong SO(3)$. Just take the quaternions and projectivize them as a 4d real vector space. I'll look at Mumford and see when he's working over $\mathbb{C}$ and when he's working over a general field... he starts out working in the category of complex analytic manifolds and then becomes more algebraic, and I may have thought he'd gone algebraic when he hadn't yet. Part of my excuse is that Wikipedia says: > Two equivalent definitions of abelian variety over a general field k are commonly in use: > * a connected and complete algebraic group over k > * a connected and projective algebraic group over k. > When the base is the field of complex numbers, these notions coincide with the previous definition. Of course I'd normally assume some idiot left out "abelian" from these definitions. But then I saw Mumford's result that says abelianness is - at least under some circumstances! - automatic, so I thought these definitions were actually correct! A further worry, which I mentioned in [[abelian variety]], is that Mumford doesn't mention the connectness assumption, at least not in any obvious way. I hate it when people say on page 43 of a 500-page book that "all varieties are assumed connected - if you don't read that page, you're toast. So I checked to see if he's doing something like that, but couldn't find it. And I wasn't smart enough to think of a finite nonabelian group that's also a zero-dimensional projective algebraic variety over $\mathbb{C}$, so I'm not sure the connectedness assumption _is_ necessary is the complex case! I'll try to straighten this stuff out. Any help would be much appreciated. This isn't part of a big research project of mine, I just want to polish things up a bit...

    Interesting issue, Todd. I don’t see any problem with your counterexample; perhaps an even easier one is $\mathbb{R}\mathrm{P}^3 \cong SO(3)$. Just take the quaternions and projectivize them as a 4d real vector space.

    I’ll look at Mumford and see when he’s working over $\mathbb{C}$ and when he’s working over a general field… he starts out working in the category of complex analytic manifolds and then becomes more algebraic, and I may have thought he’d gone algebraic when he hadn’t yet.

    Part of my excuse is that Wikipedia says:

    Two equivalent definitions of abelian variety over a general field k are commonly in use:

    • a connected and complete algebraic group over k
    • a connected and projective algebraic group over k.

    When the base is the field of complex numbers, these notions coincide with the previous definition.

    Of course I’d normally assume some idiot left out “abelian” from these definitions. But then I saw Mumford’s result that says abelianness is - at least under some circumstances! - automatic, so I thought these definitions were actually correct!

    A further worry, which I mentioned in abelian variety, is that Mumford doesn’t mention the connectness assumption, at least not in any obvious way. I hate it when people say on page 43 of a 500-page book that “all varieties are assumed connected - if you don’t read that page, you’re toast. So I checked to see if he’s doing something like that, but couldn’t find it. And I wasn’t smart enough to think of a finite nonabelian group that’s also a zero-dimensional projective algebraic variety over $\mathbb{C}$, so I’m not sure the connectedness assumption is necessary is the complex case!

    I’ll try to straighten this stuff out. Any help would be much appreciated. This isn’t part of a big research project of mine, I just want to polish things up a bit…

18. • CommentRowNumber18.
    • CommentAuthorfastlane69
    • CommentTimeAug 19th 2016
    • PermaLink
    Author: fastlane69
    Format: MarkdownItex@David > You seem to be implicitly embedding these into the circle and then thinking of the circle (which is a 1-dimensional torus) Guilty as charged! I'll be more careful to spell out "cyclic Z12" in the future, thanks. >As far as algebraic geometry goes, an abelian variety is a curve/surface/etc described by equations, and as far as I can tell are connected (EDIT: yes, see Wikipedia), in the sense that it doesn’t fall into multiple pieces – which your examples definitely do. Thanks for identifying the disconnectedness of my examples as the issue.

    @David

    You seem to be implicitly embedding these into the circle and then thinking of the circle (which is a 1-dimensional torus)

    Guilty as charged! I’ll be more careful to spell out “cyclic Z12” in the future, thanks.

    As far as algebraic geometry goes, an abelian variety is a curve/surface/etc described by equations, and as far as I can tell are connected (EDIT: yes, see Wikipedia), in the sense that it doesn’t fall into multiple pieces – which your examples definitely do.

    Thanks for identifying the disconnectedness of my examples as the issue.

19. • CommentRowNumber19.
    • CommentAuthorMike Shulman
    • CommentTimeAug 19th 2016
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexIn case anyone else is confused for the same reason I was until a moment ago: the "quaternions defined over the complex numbers" ("[biquaternions](https://en.wikipedia.org/wiki/Biquaternion)") are not a division ring, indeed they have zero-divisors, but those zero-divisors have norm 0; the elements of nonzero norm (such as those in the "unit sphere") should still be invertible, so that we do get a (nonabelian) group of unit biquaternions.

    In case anyone else is confused for the same reason I was until a moment ago: the “quaternions defined over the complex numbers” (“biquaternions”) are not a division ring, indeed they have zero-divisors, but those zero-divisors have norm 0; the elements of nonzero norm (such as those in the “unit sphere”) should still be invertible, so that we do get a (nonabelian) group of unit biquaternions.

20. • CommentRowNumber20.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 19th 2016
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexHere's a stab at a non-connected example over $k = \mathbb{C}$. The category of cocommutative coalgebras over $k$ is cocomplete (colimits are computed as in $Vect$), so we can take the left adjoint to the functor $CocommCoalg(k, -): CocommCoalg \to Set$. It takes a set $X$ to $k X$, the free vector space on $X$ with the evident coalgebra structure. The functor $X \mapsto k X$ is product-preserving. So if $G$ is a group, we have a cocommutative Hopf algebra $k G$. I think what I want is the linear dual $A = (k G)'$. This is a cogroup in the category of commutative $k$-algebras. The variety $Spec(A)$ is a group object in the category of varieties over $k$, and I guess its points correspond to elements of $G$. It is noncommutative if $G$ is, and probably under mild assumptions like finiteness of $G$, it will be complete as a variety. I might try a more careful calculation later with $G = S_3$.

    Here’s a stab at a non-connected example over $k = \mathbb{C}$.

    The category of cocommutative coalgebras over $k$ is cocomplete (colimits are computed as in $Vect$), so we can take the left adjoint to the functor $CocommCoalg(k, -): CocommCoalg \to Set$. It takes a set $X$ to $k X$, the free vector space on $X$ with the evident coalgebra structure. The functor $X \mapsto k X$ is product-preserving.

    So if $G$ is a group, we have a cocommutative Hopf algebra $k G$. I think what I want is the linear dual $A = (k G)'$. This is a cogroup in the category of commutative $k$-algebras.

    The variety $Spec(A)$ is a group object in the category of varieties over $k$, and I guess its points correspond to elements of $G$. It is noncommutative if $G$ is, and probably under mild assumptions like finiteness of $G$, it will be complete as a variety.

    I might try a more careful calculation later with $G = S_3$.

21. • CommentRowNumber21.
    • CommentAuthorJohn Baez
    • CommentTimeAug 19th 2016
    • PermaLink
    Author: John Baez
    Format: MarkdownItexI see, Todd: so you're getting a finite group as an affine rather than projective variety, but then you're saying "hey, but it's complete, so Mumford's result, whatever it is, should still apply". Let me spend 15 more minutes trying to figure out exactly Mumford proved.

    I see, Todd: so you’re getting a finite group as an affine rather than projective variety, but then you’re saying “hey, but it’s complete, so Mumford’s result, whatever it is, should still apply”.

    Let me spend 15 more minutes trying to figure out exactly Mumford proved.

22. • CommentRowNumber22.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 19th 2016
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexYeah, it's analogous to saying that if you have a compact space, then taking its completion in an ambient space adds nothing new. Here we take the projective completion of a finite affine variety, and get nothing new.

    Yeah, it’s analogous to saying that if you have a compact space, then taking its completion in an ambient space adds nothing new. Here we take the projective completion of a finite affine variety, and get nothing new.

23. • CommentRowNumber23.
    • CommentAuthorJohn Baez
    • CommentTimeAug 19th 2016
    • (edited Aug 19th 2016)
    • PermaLink
    Author: John Baez
    Format: MarkdownItexOkay, so Mumford is assuming his field is algebraically closed! That kills off the counterexamples over $\mathbb{R}$. Sorry, Todd - thanks for catching that. I'll fix that in the $n$Lab. In the beginning of Chapter 2 of _Abelian Varieties_ he says: > We now turn to the study of abelian varieties over an arbitrary algebraically closed field $k$. > **Definition.** An abelian variety $X$ is a complete algebraic variety${}^\dagger$ over $k$ with a group law $m: X \times X \to X$ such that $m$ and the inverse map are both morphisms of varieties. Note, no commutative law assumed; he derives that later: > We will show that $X$ is a commutative and divisible group. So in particular he somehow manages to rule out finite abelian groups. But what's that footnote? > ${}^\dagger$ This means, in particular, that it is irreducible. Hmm. I don't know this stuff very well: does this imply it's connected? He actually gives two proofs that $X$ is abelian. In the middle of the first he says: > Since... and $X$ is complete and connected... Where did the connectedness come from??? I checked all pages leading up to this page (which is, luckily, only page 41), and he never says anything about some additional implicit assumption of connectedness. So I have to hope that complete algebraic varieties over algebraically closed fields are automatically connected for some reason. I can easily imagine that irreducibility implies connectedness. I don't know why he says completeness implies irreducibility, but I'll take his word for it. (I'll also admit I don't know the definition of connectedness of varieties over arbitrary fields, though I can make one up, like saying the only idempotents in the field of functions are 0 and 1.)

    Okay, so Mumford is assuming his field is algebraically closed! That kills off the counterexamples over $\mathbb{R}$. Sorry, Todd - thanks for catching that. I’ll fix that in the $n$Lab.

    In the beginning of Chapter 2 of Abelian Varieties he says:

    We now turn to the study of abelian varieties over an arbitrary algebraically closed field $k$.

    Definition. An abelian variety $X$ is a complete algebraic variety${}^\dagger$ over $k$ with a group law $m: X \times X \to X$ such that $m$ and the inverse map are both morphisms of varieties.

    Note, no commutative law assumed; he derives that later:

    We will show that $X$ is a commutative and divisible group.

    So in particular he somehow manages to rule out finite abelian groups. But what’s that footnote?

    ${}^\dagger$ This means, in particular, that it is irreducible.

    Hmm. I don’t know this stuff very well: does this imply it’s connected? He actually gives two proofs that $X$ is abelian. In the middle of the first he says:

    Since… and $X$ is complete and connected…

    Where did the connectedness come from??? I checked all pages leading up to this page (which is, luckily, only page 41), and he never says anything about some additional implicit assumption of connectedness. So I have to hope that complete algebraic varieties over algebraically closed fields are automatically connected for some reason. I can easily imagine that irreducibility implies connectedness. I don’t know why he says completeness implies irreducibility, but I’ll take his word for it.

    (I’ll also admit I don’t know the definition of connectedness of varieties over arbitrary fields, though I can make one up, like saying the only idempotents in the field of functions are 0 and 1.)

24. • CommentRowNumber24.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 19th 2016
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexOh... an affine *variety* is irreducible by definition (Wikipedia), i.e., is the zero locus of a *prime* ideal. If we remove the primeness condition, we speak instead of an algebraic set. So that takes care of my last proposed counterexample. It's starting to make more sense now.

    Oh… an affine variety is irreducible by definition (Wikipedia), i.e., is the zero locus of a prime ideal. If we remove the primeness condition, we speak instead of an algebraic set. So that takes care of my last proposed counterexample. It’s starting to make more sense now.

25. • CommentRowNumber25.
    • CommentAuthorJohn Baez
    • CommentTimeAug 19th 2016
    • PermaLink
    Author: John Baez
    Format: MarkdownItexOh, okay! So affine varieties are all irreducible by definition - who knew? Does this in turn imply they're connected? I'm glad I'm not the only guy here who learned algebraic geometry on the street. <img src = "http://imgc-cn.artprintimages.com/images/P-473-488-90/60/6005/GXFB100Z/posters/roz-chast-the-kid-who-learned-about-math-on-the-street-cartoon.jpg" alt = ""/>

    Oh, okay! So affine varieties are all irreducible by definition - who knew? Does this in turn imply they’re connected?

    I’m glad I’m not the only guy here who learned algebraic geometry on the street.

    

26. • CommentRowNumber26.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 19th 2016
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexIt might imply connectedness over $\mathbb{C}$. Not over $\mathbb{R}$ though. For example, the pictures you see [here](https://en.wikipedia.org/wiki/Elliptic_curve) are of the real points of genuine elliptic curves, many of which are not connected. However, the complex variety defined by the same equation *is* connected. (I might be able to cook up a simpler example later, but time for bed.)

    It might imply connectedness over $\mathbb{C}$. Not over $\mathbb{R}$ though. For example, the pictures you see here are of the real points of genuine elliptic curves, many of which are not connected. However, the complex variety defined by the same equation is connected. (I might be able to cook up a simpler example later, but time for bed.)

27. • CommentRowNumber27.
    • CommentAuthorMike Shulman
    • CommentTimeAug 19th 2016
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexI wish people would stop including conditions like "connected" and "irreducible" in the basic definitions of their subject. The basic objects ought to form a [[dichotomy between nice objects and nice categories|nice category]].

    I wish people would stop including conditions like “connected” and “irreducible” in the basic definitions of their subject. The basic objects ought to form a nice category.

28. • CommentRowNumber28.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 19th 2016
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexMike, I agree. I guess never underestimate the power of inertia in teaching the foundations of a subject, not to mention in not taking categories seriously.

    Mike, I agree. I guess never underestimate the power of inertia in teaching the foundations of a subject, not to mention in not taking categories seriously.

29. • CommentRowNumber29.
    • CommentAuthorJohn Baez
    • CommentTimeAug 21st 2016
    • (edited Aug 21st 2016)
    • PermaLink
    Author: John Baez
    Format: MarkdownItexMike wrote: > I wish people would stop including conditions like “connected” and “irreducible” in the basic definitions of their subject. There's no point saying it here - you're preaching to the converted. You should organize protests at graduate math courses.

    Mike wrote:

    I wish people would stop including conditions like “connected” and “irreducible” in the basic definitions of their subject.

    There’s no point saying it here - you’re preaching to the converted. You should organize protests at graduate math courses.