# Interview with a Mathematical Physicist: John Baez Part 1 - Comments

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Brilliant Insight into your life John! Are you planning to returning to the "Centre of Quantum Technologies" this year? Also I must imagine you have the most interesting dinner discussions with your wife. Philosophy and Physics :)

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Brilliant Insight into your life John! Are you planning to returning to the "Centre of Quantum Technologies" this year? Also I must imagine you have the most interesting dinner discussions with your wife. Philosophy and Physics :)
Thanks for giving me the chance to think about where I've been and where I'm going!

I'll go back to the Centre for Quantum Technologies again at the end of June. My wife has been working at the Philosophy Department at NUS while I've been working there. We think this will be our last summer working in Singapore: we've had a good run of it, but we're ready to try something new. Frankly I'd be happy to stay home and tend to our garden---I always get nervous about the plants when I'm gone during the hot summers in Riverside! But I suspect we'll probably go to Europe next summer.

Thanks for giving me the chance to think about where I've been and where I'm going!

I'll go back to the Centre for Quantum Technologies again at the end of June. My wife has been working at the Philosophy Department at NUS while I've been working there. We think this will be our last summer working in Singapore: we've had a good run of it, but we're ready to try something new. Frankly I'd be happy to stay home and tend to our garden---I always get nervous about the plants when I'm gone during the hot summers in Riverside! But I suspect we'll probably go to Europe next summer.
Do you by any chance know Helmer Aslaksan? I would think so.

This article uses an E8 projection which I introduced to Wikipedia in Feb of 2010 here. Technically, it is E8 projected to the E6 Coxeter plane.

The projection uses X Y basis vectors of:
X = {-Sqrt[3] + 1, 0, 1, 1, 0, 0, 0, 0};
Y = {0, Sqrt[3] - 1, -1, 1, 0, 0, 0, 0};

Resulting in vertex overlaps of:
24 Red with 1 overlap
24 Orange each with 8 overlaps (192 vertices)
1 Yellow with 24 overlaps (24 vertices)

This was subsequently recreated in the current article and 4_21 E8 WP page by Tom Ruen, source for the article.

After doing this for a few example symmetries, Tom took my idea of projecting higher dimensional objects to the 2D (and 3D) symmetries of lower dimensional subgroups - and ran with it in 2D - producing a ton of visualizations across WP. :-)

http://theoryofeverything.org/theToE/2016/03/16/e8-in-e6-petrie-projection/

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Do you by any chance know Helmer Aslaksan? I would think so.
No, I don't know him. He seems like a cool guy, but I haven't been hanging out in the math department at NUS.

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This article uses an E8 projection which I introduced to Wikipedia in Feb of 2010 here.
Cool! Maybe you mean E6 projection? I deliberately chose an image that was in the public domain so I wouldn't have to give attributions, but thanks for helping come up with it!

The 72 vertex and 720 edge E6 (as a subgroup of E8) in the E6 Coxeter plane (a Petrie projection) produces vertex overlaps of:
24 Yellow with 1 overlap
24 Orange each with 2 overlaps (48 vertices)
with none at the origin.

(which I have now included here:
http://theoryofeverything.org/theToE/2016/03/16/e8-in-e6-petrie-projection/)

See this work by Stembridge on this as well: http://www.math.lsa.umich.edu/~jrs/coxplane.html

Taking all 240 of E8 vertices and 6720 edges produces a similar result but different (as you used in the article).

BTW - great article!

oh, and a minor point... it is my understanding that citation (author name and link if online) is still required to WP WikiMedia commons content. (of course, in this case the work was Tom Ruen's - he simply took my basis vectors and produced the image using his own source code (rather than using the Mathematica tool I use and offer in the public domain).

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oh, and a minor point... it is my understanding that citation (author name and link if online) is still required to WP WikiMedia commons content.
I believe someone puts something in the public domain, that means anyone can do anything they want with it. If you read the page on which this image appears on Wikicommons you'll see it says:

I, the copyright holder of this work, release this work into the public domain. This applies worldwide.
In some countries this may not be legally possible; if so:

I grant anyone the right to use this work for any purpose, without any conditions, unless such conditions are required by law.
One reason I like Tom Ruen's work so much is that he puts it into the public domain, thus freeing it up for worldwide use without any need for attribution. It's a gift to the universe.

But of course, if Mr. X tries to put Mr. Y's copyrighted work into the public domain, Mr. Y can argue with that.

I stand corrected - it seems Tom does (at least for some images) change the default file upload CC Share-Alike Commons to "Public Domain".

In checking though, he does have others that have the default (as, AFAIK, mine do) which does require attribution. So I guess, we need to double check each image.

Don't get me wrong, I am ok w/putting my stuff in public domain w/o attribution requirements. In using the defaults I assumed proper etiquette was to cite WP sources as a matter of course.

e.g. https://commons.wikimedia.org/wiki/File:Flower_of_life_triangular_11547-arccircle.svg

Licensing

I, the copyright holder of this work, hereby publish it under the following license:

You are free:
• to share – to copy, distribute and transmit the work
• to remix – to adapt the work
Under the following conditions:
• attribution – You must attribute the work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work).
• share alike – If you alter, transform, or build upon this work, you may distribute the resulting work only under the same or similar license to this one.

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By the way, thanks for explaining how many vertices overlap in this 2d projection of the E8 root polytope!

ShayanJ
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Such number systems only exist in 1, 2, 4, and 8 dimensions
I remember reading somewhere that this is somehow because of the fact that "every cow must have at least one cowlick", but there wasn't much explanation! Can anybody give a clue?

Also, interesting interview, thanks!

I remember reading somewhere that this is somehow because of the fact that "every cow must have at least one cowlick", but there wasn't much explanation! Can anybody give a clue?

Also, interesting interview, thanks!
It has nothing to do with it. The "cow" theorem is about poles in odd dimensional spaces.

I remember reading somewhere that this is somehow because of the fact that "every cow must have at least one cowlick", but there wasn't much explanation! Can anybody give a clue?

Also, interesting interview, thanks!
It has nothing to do with it. The "cow" theorem is about poles in odd dimensional spaces.
I'm not sure, but I think Shyan's question was an attempt at humor (I laughed anyway).

Yet, if he was asking a serious question, the reason is that John was describing the conditions for "normed division algebras".

The 1,2,4 and 8 dimensions are due to the 2^n nature of the Cayley-Dickson construction of the algebras.

Add to this, given the reference to "division" algebra, it ends at dimension 8 with octonions because the next level up (n=4), the 16 dimensional sedenions, contain "zero divisors" which prevents division in those cases. For my list of those zeros (and a cool animation), see:
https://en.wikipedia.org/wiki/Sedenion

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atyy
Hmm, biochemistry. How about neurobiology? There seem to be networks there too.

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I remember reading somewhere that this is somehow because of the fact that "every cow must have at least one cowlick", but there wasn't much explanation! Can anybody give a clue?
Here's the relation. A sphere in n-dimensional space can have least one continuous nowhere vanishing vector field if and only if n = 2,4,6,8,... A sphere in n-dimensional space can have (n-1) linearly independent continuous vector fields if n = 1, 2, 4, or 8.

People know, for a sphere of any dimension, the maximum number of linearly independent continuous vector fields on it. See:
These results subsume everything I just said, and more.

Also, interesting interview, thanks!
You're welcome!

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Hmm, biochemistry. How about neurobiology? There seem to be networks there too.
My concept of biology is so broad that it includes biochemistry, neurobiology, ecology and more. The network formalisms I'm developing are so general that they should have some relevance to all of these topics... though of course it'd take expertise to develop any one particular application to the point of doing something useful!

I hope I'm not beating a dead cow here, but....

The octonions are a number system where you can add, subtract, multiply and divide. Such number systems only exist in 1, 2, 4, and 8 dimensions
and
...that this is somehow because of the fact that "every cow must have at least one cowlick"...
and
Here's the relation. ... A sphere in n-dimensional space can have (n-1) linearly independent continuous vector fields if n = 1, 2, 4, or 8.
The "relation" is as John pointed out in the WP reference, yet the "cause and effect" seem to be reversed in Shyan's comment. That is: the 1,2,4 and 8 limit on the dimensionality of numbers systems where you can add, subtract, multiply and divide is not "because of" the number of linearly independent continuous vector fields (or their number of hairy ball cowlicks).

It is the octonion limit on the number of dimensions (n=8) for normed division algebras (due to sedenions having zero divisors and the Cayley-Dickson construction) that constrains n<9. The number of linearly independent continuous vector fields within the modulo 8 periodicity of Clifford Algebras are related yet dependent rather than being a cause of any lack of "normed division" capability.

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I think it's awesome that we have some prominent scientists and mathematicians here on PF.