Skip to the Main Content

Note:These pages make extensive use of the latest XHTML and CSS Standards. They ought to look great in any standards-compliant modern browser. Unfortunately, they will probably look horrible in older browsers, like Netscape 4.x and IE 4.x. Moreover, many posts use MathML, which is, currently only supported in Mozilla. My best suggestion (and you will thank me when surfing an ever-increasing number of sites on the web which have been crafted to use the new standards) is to upgrade to the latest version of your browser. If that's not possible, consider moving to the Standards-compliant and open-source Mozilla browser.

January 27, 2012

G2 and Spin(8) Triality

Oscar Chacaltana, Yuji Tachikawa and I are deep in the weeds of nilpotent orbits. One of the things we had to study were the nilpotent orbits of 𝔤 2\mathfrak{g}_2, and how they sit in 𝔰𝔬(8)\mathfrak{so}(8). Understanding the answer involves an explicit description of Spin(8)Spin(8) triality, which I thought was kinda cute. Few people will care about the nilpotent orbits, but the bit about triality and G 2G_2 might be of some independent interest. So here it is.

Spin(8)Spin(8) has a triality symmetry (an outer autmomorphism of the Lie algebra), which permutes the three 8-dimensional irreducible representations: 8 v8_v, 8 s8_s, and 8 c8_c. 𝔤 2𝔰𝔬(8)\mathfrak{g}_2\subset \mathfrak{so}(8) is the invariant subalgebra. (I’ll conveniently pass back and forth between the complex form of the Lie algebra and the compact real form of the group, as both are of interest to us.) What I want to do is describe that triality symmetry very explicitly and, thereby, the realization of 𝔤 2\mathfrak{g}_2.

Note that Spin(8)Spin(8) contains an (SU(2) 4)/ 2({SU(2)}^4)/\mathbb{Z}_2 subgroup, under which the adjoint decomposes as 28=(3,1,1,1)+(1,3,1,1)+(1,1,3,1)+(1,1,1,3)+(2,2,2,2) 28 = (3,1,1,1)+(1,3,1,1)+(1,1,3,1)+(1,1,1,3)+(2,2,2,2) Under this decomposition, the action of triality is easy to describe: pick one of the 𝔰𝔩(2)\mathfrak{sl}(2) subalgebras to hold fixed, and consider all permutations of the other three (supplemented by the obvious action on the (2,2,2,2)(2,2,2,2)).

That’s triality. Looked at this way, it seems absurdly simple. The above description gives a perfectly concrete action of triality, as permutations of the generators. And we can push a little harder, and really understand 𝔤 2\mathfrak{g}_2, this way.

The subalgebra, invariant under the S 3S_3 permutations, is 𝔤 2𝔰𝔬(8)\mathfrak{g}_2\subset \mathfrak{so}(8), under which 28=14+7V 28 = 14 + 7 \otimes V where VV is the 2-dimensional irreducible representation of S 3S_3. In terms of our previous decomposition, G 2(SU(2)×SU(2) D)/ 2 G_2 \supset (SU(2)\times {SU(2)}_D)/\mathbb{Z}_2 where the first SU(2)SU(2) is the one you kept fixed, and SU(2) D{SU(2)}_D is the diagonal SU(2)SU(2) of the three which are permuted by triality. Under this embedding, 14 =(3,1)+(1,3)+(2,4) 7 =(1,3)+(2,2) \begin{split} 14 &= (3,1)+(1,3)+(2,4)\\ 7 &= (1,3) + (2,2) \end{split}

An explicit basis of antisymmetric 8×88\times 8 matrices which give this 𝔤 2\mathfrak{g}_2 subalgebra is as follows. First, we embed 𝔰𝔩(2) 4{\mathfrak{sl}(2)}^4, by taking the 8×88\times 8 matrix to be block-diagonal, with 4×44\times 4 blocks containing 𝔰𝔩(2) 2{\mathfrak{sl}(2)}^2, as H L =σ 2𝟙 X L =12(σ 3+iσ 1)σ 2 Y L =12(σ 3iσ 1)σ 2=X L ,H R =𝟙σ 2 X R =12σ 2(σ 3+iσ 1) Y R =12σ 2(σ 3iσ 1)=X R \begin{matrix} \begin{split} H_L &= \sigma_2 \otimes \mathbb{1}\\ X_L &= \tfrac{1}{2} (\sigma_3+i\sigma_1) \otimes \sigma_2\\ Y_L &= \tfrac{1}{2} (\sigma_3-i\sigma_1) \otimes \sigma_2 = X_L^\dagger \end{split}&,\qquad\qquad \begin{split} H_R &= \mathbb{1} \otimes \sigma_2\\ X_R &= \tfrac{1}{2} \sigma_2 \otimes (\sigma_3+i\sigma_1)\\ Y_R &= \tfrac{1}{2} \sigma_2 \otimes (\sigma_3-i\sigma_1) = X_R^\dagger \end{split} \end{matrix} where we’ve chosen the normalization conventions [X,Y] =H [H,X] =2X [H,Y] =2Y \begin{split} [X,Y]&=H\\ [H,X]&=2X\\ [H,Y]&=-2Y \end{split}

We pick one of these (the 𝔰𝔩(2) L{\mathfrak{sl}(2)}_L in the upper left-hand block) to hold fixed, and embed our second 𝔰𝔩(2)\mathfrak{sl}(2) diagonally in the other three: H 1 =12(𝟙+σ 3)σ 21 X 1 =14(𝟙+σ 3)(σ 3+iσ 1)σ 2 Y 1 =X 1 H 2 =12(𝟙σ 3)σ 2𝟙+𝟙𝟙σ 2 X 2 =14(𝟙σ 3)(σ 3+iσ 1)σ 2+121σ 2(σ 3+iσ 1) Y 2 =X 2 \begin{split} H_1 &= \tfrac{1}{2} (\mathbb{1}+\sigma_3)\otimes \sigma_2 \otimes 1\\ X_1 &= \tfrac{1}{4} (\mathbb{1}+\sigma_3)\otimes (\sigma_3+i\sigma_1)\otimes\sigma_2\\ Y_1 &= X_1^\dagger\\ H_2 &= \tfrac{1}{2} (\mathbb{1}-\sigma_3)\otimes \sigma_2 \otimes \mathbb{1} + \mathbb{1}\otimes \mathbb{1} \otimes \sigma_2\\ X_2 &= \tfrac{1}{4} (\mathbb{1}-\sigma_3)\otimes (\sigma_3+i\sigma_1)\otimes \sigma_2 + \tfrac{1}{2} 1\otimes \sigma_2\otimes (\sigma_3+i\sigma_1)\\ Y_2 &= X_2^\dagger \end{split}

The highest weight of the (2,4)(2,4) is S 1,3=14σ 2(σ 3+iσ 1)(σ 3+iσ 1) S_{1,3} = \tfrac{1}{4} \sigma_2\otimes (\sigma_3+i\sigma_1)\otimes (\sigma_3+i\sigma_1) The remaining ones, e.g., S 1,3=[Y 1,S 1,3]S_{-1,3} = [Y_1, S_{1,3}], are obtained by acting with the lowering operators, Y 1,2Y_{1,2}. With this choice of Cartan, the simple roots of 𝔤 2\mathfrak{g}_2 correspond to X 2X_2 (short root) and S 1,3S_{1,-3} (long root).

Layer 1 X 2 X_2 S 1 , 3 S_{1,-3} \begin{svg}<svg width="96" height="44" xmlns="http://www.w3.org/2000/svg" xmlns:se="http://svg-edit.googlecode.com"> <g> <title>Layer 1</title> <foreignObject x="3.164062" y="0.125" id="svg_45629_8" font-size="16" width="20" height="20"> <math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"> <semantics> <mrow> <msub> <mi>X</mi> <mn>2</mn> </msub> </mrow> <annotation encoding="application/x-tex">X_2</annotation> </semantics> </math> </foreignObject> <foreignObject x="59.039062" y="0" id="svg_45629_9" font-size="16" width="36" height="22"> <math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"> <semantics> <mrow> <msub> <mi>S</mi> <mrow> <mn>1</mn> <mo>,</mo> <mo lspace="verythinmathspace" rspace="0em">&#x2212;</mo> <mn>3</mn> </mrow> </msub> </mrow> <annotation encoding="application/x-tex">S_{1,-3}</annotation> </semantics> </math> </foreignObject> <line fill="none" stroke="#000000" stroke-width="2" x1="17.483687" y1="26.794221" x2="57.969444" y2="26.794221" id="svg_45629_1"/> <line fill="none" stroke="#000000" stroke-width="2" x1="17.483687" y1="30.442583" x2="57.969444" y2="30.442583" id="svg_45629_2"/> <line fill="none" stroke="#000000" stroke-width="2" x1="17.483687" y1="34.090945" x2="57.969444" y2="34.090945" id="svg_45629_3"/> <circle fill="#ffffff" stroke="#000000" stroke-width="2" cx="9.799556" cy="30.442583" r="7.798828" id="svg_45629_4"/> <circle fill="#000000" stroke="#000000" stroke-width="2" cx="65.653574" cy="30.442583" r="7.798828" id="svg_45629_5"/> <path fill="none" stroke="#000000" stroke-width="3" stroke-linecap="round" d="m43.90065,19.679909l-12.044514,10.853888l12.145721,11.036291" id="svg_45629_7"/> </g></svg>\end{svg}

This 8-dimensional representation of G 2G_2, as it’s reducible, is not the most convenient one for studying the representation theory of G 2G_2. But it’s tailor-made for our purpose, which is understanding the embedding in Spin(8)Spin(8). With an explicit embedding in hand, we can manufacture a distinguished triple, (H,X,Y)(H,X,Y) for each nilpotent orbit of 𝔤 2\mathfrak{g}_2, and see how it sits in 𝔰𝔬(8)\mathfrak{so}(8). But that, probably, holds little interest for the general reader, so I’ll end here.

Posted by distler at January 27, 2012 2:09 PM

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2491

11 Comments & 1 Trackback

Re: G2 and Spin(8) Triality

“But that, probably, holds little interest for the general reader, so I’ll end here.”

But what’s *your* interest in the nilpotent orbits? Is there a physical model where they matter?

Posted by: Michael on January 28, 2012 10:10 AM | Permalink | Reply to this

Re: G2 and Spin(8) Triality

I suppose you could wait for the paper.

Or you could look at our previous papers for clues.

Posted by: Jacques Distler on January 28, 2012 12:29 PM | Permalink | PGP Sig | Reply to this

Re: G2 and Spin(8) Triality

Nice! Small typo: it looks like you defined X RX_R twice. I think the second one should be Y RY_R. Sorry that I can’t point to an equation since the equations are not numbered.

By the way, in my Firefox 9.0.1, the alignment of the equations is a bit off. There is more space before the equal sign than after.

Posted by: Sidious Lord on January 29, 2012 3:01 AM | Permalink | Reply to this

Re: G2 and Spin(8) Triality

Small typo: it looks like you defined X RX_R twice. I think the second one should be Y RY_R.

Fixed. Thanks!

By the way, in my Firefox 9.0.1, the alignment of the equations is a bit off. There is more space before the equal sign than after.

Yes, that’s a known bug (lack of/broken implementation of the columnspacing attribute on the mtable element).

Posted by: Jacques Distler on January 29, 2012 3:11 PM | Permalink | PGP Sig | Reply to this

Re: G2 and Spin(8) Triality

Dear Prof Distler,

I am a graduate student and intend to work on theoretical high energy physics. I want to learn techniques in the fields of algebraic geometry and topology, so that I am able to broadly understand current research papers in the field. Can you suggest me some pedagogical and introductory review articles/books suited for me? I have done a semester course each in quantum field theory and general relativity.

Thank you,
Prashant

Posted by: Prashant on January 31, 2012 2:45 PM | Permalink | Reply to this

Re: G2 and Spin(8) Triality

It’s hard to know, in advance, which bits of mathematics will be useful in your work. But Griffiths and Harris is still the best place to find an introduction to Algebraic Geometry.

Posted by: Jacques Distler on January 31, 2012 2:55 PM | Permalink | PGP Sig | Reply to this

Re: G2 and Spin(8) Triality

I find the description via root systems to be a bit easier: triality acts on the roots in Spin(8), and each long root in G2 comes from a fixed root for triality, while each short root in G2 comes from an orbit of 3 roots for triality.

You can make this more precise in terms of automorphisms of Spin(8)/so(8) and a basis for a root system of type D4.

Of course you do point this out, but there is a very natural embedding of G2 in Spin(7). The nilpotent orbits in G2 and representatives/distinguished triples are (of course) very well understood. There’s a paper by Kraft on the geometry, though that may not be what you’re interested in.

Posted by: Paul Levy on March 5, 2012 4:25 PM | Permalink | Reply to this

Re: G2 and Spin(8) Triality

Perhaps you can write a post on Nina Arkani-Hamed’s talk from this afternoon. That would be greatly appreciated!

Posted by: uzbecker on March 7, 2012 10:01 PM | Permalink | Reply to this

Re: G2 and Spin(8) Triality

I mean Nima.

Posted by: uzbecker on March 8, 2012 12:06 AM | Permalink | Reply to this
Read the post Spring Break
Weblog: Musings
Excerpt: Stuck in College Station
Tracked: March 14, 2012 1:04 AM

Re: G2 and Spin(8) Triality

There is a nice description of triality and G 2G_2 using octonions. The group G 2G_2 is automorphism group of the octonionic algebra and elements of Spin(8)Spin(8)can be realized as explicit (real) endomorphisms of 𝕆 2\mathbb{O}^2 and the triality is manifested by explicit embedding of Spin(8)Spin(8) into SO(8)×SO(8)×SO(8)SO(8)\times SO(8) \times SO(8). See unpublished note by Bryant for details. (http://www.math.duke.edu/~bryant/Spinors.pdf)

These things are also presented in more generality in book by Sprigner and Veldkamp. Also there were some exposition articles on triality by Varadarajan and Sapio in Expositiones Mathematicae.

Posted by: Vit Tucek on May 26, 2012 9:23 AM | Permalink | Reply to this

Re: G2 and Spin(8) Triality

Another explicit embedding can be found in this paper, specifically in Appendix C. We split so(8) = so(7) + 7 = g_2 + 7 + 7, using the fact that g_2 preserves the octonion structure constants.

Posted by: Craig on August 7, 2012 4:25 PM | Permalink | Reply to this

Post a New Comment