Why is it necessary to have 11 dimensions in M Theory?

[B936]

I do apologize if this is the wrong place to post this.

Why is it necessary to have 11 dimensions in M Theory? I've been at it for days emailing universities, Twitter, etc. No one has attempted to give me an answer and it's frustrating me.

I just want to know what the purpose of each of the dimensions of M Theory are and why it couldn't work with only 10 dimensions.

Thanks

 

[Demystifier]

In physical theories it is very important to have symmetries. It is easy to construct classical theories with symmetries, but quantum effects often destroy symmetries of the classical theory. In general, it may be very difficult to construct a quantum theory which will save all the symmetries of the classical theory.

The string theory is not an exception. In turns out that quantum string theory always destroys the symmetries of the classical string theory, except in one special case: when the number of dimensions is 10. That's why string theory works only in 10 dimensions. The most surprising fact about that is that there is any number of dimensions at all for which symmetries are not destroyed. Before calculation, one would intuitively expect that the symmetries would be destroyed in any number of dimensions.

But string theory is not one theory. There are 5 different string theories, each requiring 10 dimensions. On the other hand, string theory is supposed to be fundamental theory. How can there be 5 different fundamental theories? A possible answer is that neither of them is really fundamental. Instead, they can all be special cases of a more fundamental theory, called M-theory. The M-theory is therefore more general than string theory. But since it is more general, it turns out that it must have more dimensions than string theory. The smallest possible number bigger than 10 is 11, and it seems that 11 is a sufficient number to fit all 5 string theories into one bigger theory.

 

[Ashwin2012]

But since it is more general, it turns out that it must have more dimensions than string theory. The smallest possible number bigger than 10 is 11, and it seems that 11 is a sufficient number to fit all 5 string theories into one bigger theory.

In fact S-duality relation implies , in the UV region of 10 -d type IIA theory (one of the dimension) decompactifies to give us the structure of M- theory.And such a parallel was found in the supergravity theories as well.
Hence , N=1 , 11D SUGRA is considered as the low energy limit of M theory.

 

[jeffery_winkle]

The reason comes from the 3-Psi rule which John Baez talks about here.

http://math.ucr.edu/home/baez/week279.html

The four normed division algebras are

Real numbers - dimension 1
Complex numbers - dimension 2
Quaternions - dimension 4
Octonions - dimension 8

The one relevant to string theory are the octonions which are 8 dimensional. A one dimensional string sweeps out a 2D worldsheet. 8 + 2 = 10. Therefore string theory is only consistent in 10 dimensions. M-theory uses 2-branes as the fundamental entity. A 2-brane sweeps out a 3D worldvolume. 8 + 3 = 11. Therefore M-theory is only consistent in 11 dimensions.

 

[Demystifier]

The reason comes from the 3-Psi rule which John Baez talks about here.

http://math.ucr.edu/home/baez/week279.html

The four normed division algebras are

Real numbers - dimension 1
Complex numbers - dimension 2
Quaternions - dimension 4
Octonions - dimension 8

The one relevant to string theory are the octonions which are 8 dimensional. A one dimensional string sweeps out a 2D worldsheet. 8 + 2 = 10. Therefore string theory is only consistent in 10 dimensions. M-theory uses 2-branes as the fundamental entity. A 2-brane sweeps out a 3D worldvolume. 8 + 3 = 11. Therefore M-theory is only consistent in 11 dimensions.

Let me pretend that I buy it. How then one would explain 26 dimensions of bosonic string theory?

 

[jeffery_winkle]

Well, in that case, one would read John Baez's other article explaining why the number 24 is one of his favorite numbers!

http://math.ucr.edu/home/baez/numbers/24.pdf

 

[Demystifier]

That's great! But I am still not sure should I take it seriously or as a very profound joke.

Perhaps Baez can also explain why we see precisely 4 dimensions? Let me guess, that's because quaternions live in 4 dimensions.

 

[arivero]

That's great! But I am still not sure should I take it seriously or as a very profound joke.

Perhaps Baez can also explain why we see precisely 4 dimensions? Let me guess, that's because quaternions live in 4 dimensions.

No, the Susy <==> triality <====> Division Algebras thing is not a joke, as Baez points out, it is well known

6) Taichiro Kugo and Paul Townsend, Supersymmetry and the division algebras, Nucl. Phys. B 221 (1983), 357-380. Also available at http://ccdb4fs.kek.jp/cgi-bin/img_index?198301032
7) J. M. Evans, Supersymmetric Yang-Mills theories and division algebras, Nucl. Phys. B 298 (1988), 92-108. Also available at http://ccdb4fs.kek.jp/cgi-bin/img_index?8801412

The industry has been heavely exploited by Duff in the "Brane Scan".

As for why the collapse is to 4+7, the only suggestion I know is that it comes from the number of indices of the bosonic parnert of the graviton, an antisymmetric tensor $A_{NMP}$.

 

[samalkhaiat]

I do apologize if this is the wrong place to post this.

Why is it necessary to have 11 dimensions in M Theory? I've been at it for days emailing universities, Twitter, etc. No one has attempted to give me an answer and it's frustrating me.

I just want to know what the purpose of each of the dimensions of M Theory are and why it couldn't work with only 10 dimensions.

Thanks

Given a Lie group $G$ and its maximal subgroup $H$, then the coset space $G/H$ is the lowest-dimensional manifold which can have $G$ as a symmetry group: $\mbox{dim}(G/H) = \mbox{dim}(G) - \mbox{dim}(H)$. Now consider the following examples:
If $G = ISO(1,3)$, the 10-dimensional Poincare group, $H = SO(1,3)$ is the 6-dimensional Lorentz group. We can identify the 4-dimensional coset malifold $ISO(1,3) /SO(1,3) = M^{4}$ with the Poincare’ symmetric 4D Minkowski space-time.
If $G = SU(2)$: $\mbox{dim}(SU(2) ) = 3$, $H = U(1)$: $\mbox{dim}(U(1)) =1$, then $SU(2)/U(1) = S^{2}$, where $S^{2}$ is the $SU(2)$-symmetric 2-dimensional sphere.
Now, take $G = SU(3)$: $\mbox{dim}(SU(3)) = 8$, its maximal subgroup is the 4-dimensional group $H = SU(2) \times U(1)$, and $$SU(3) / SU(2) \times U(1) = \mbox{CP}^{2} ,$$ is the $SU(3)$-symmetric 4-dimensional (complex) projective space. We also know that the circle $S^{1}$ is one dimensional, $U(1)$-symmetric space. Thus, the space $\mathcal{V}^{7} = \mbox{CP}^{2} \times S^{2} \times S^{1}$ has 7 dimensions and admits the symmetry group of the SM, i.e. $SU(3) \times SU(2) \times U(1)$. Clearly, $M^{4} \times \mathcal{V}^{7}$ is 11-dimensional space symmetric under Poincare' and the gauge group of the SM: $ISO(1,3) \times SU(3) \times SU(2) \times U(1)$.
So, if you want to formulate a Kaluza-Klein type theory in which $su(3) \times su(2) \times u(1)$-valued gauge fields arise as components of the metric tensor, $g_{AB}(x)$, in more than 4 (non-compact) space-time dimensions, you must have at least 7 extra dimensions, i.e., $A, B = 1, 2, \cdots , 11$. Thus D = 11 is the minimum number with which you can obtain the gauge fields of the gauge group $SU(3) \times SU(2) \times U(1)$ by Kaluza-Klein method. Since, we have never been able to formulate a consistent field theory with gravity coupled to massless spin > 2 particles. And, since $D >11$ supergravity contains such massless, spin>2 particle, we conclude that 11 dimensions is the maximum for consistent supergravity. It seems like a remarkable coincidence that D=11, which is the minimum number required by K-K procedure, is the maximum number required by consistent supergravity.

Sam

 

[Demystifier]

11 dimensions is the maximum for consistent supergravity

Then what about F-theory, which lives in 12 dimensions?

 

[MathematicalPhysicist]

@Demystifier perhaps he refers to 8+3 spatial dimension where 8 is the maximal spatial dimensions, the added dimension in F theory is a temporal dimension.

 

[samalkhaiat]

Then what about F-theory, which lives in 12 dimensions?

What is that has to do with my post? I was making a group theoretical justification for the ''why $D =11$ in M-theory'', and the fact that $D_{min} = 11$ M-theory has $D_{max}=11, \ N =1$ supergravity as its low energy limit. What is that has to do with F-theory? Do you even know the metric signature in F-theory?
With signature $(D -1,1)$ and for $D \geq 12$, the supersymmetric relation $N_{B} = N_{F}$ can not be satisfied [1]. Soon after Nahm’s paper, Cremmer, Julia and Scherk [2] showed that supergravity not only permits up to 7 extra dimensions but takes its simplest form in eleven dimensions. Furthermore, without the need for any assumptions about higher spin, Duff & Lu [3] showed that $D=11$ emerges naturally as the maximum spacetime dimension admitting supersymmetric extended objects (p-brane), i.e. the “brane-scan” stops at $D = 11$.


[1] W. Nahm, Nucl. Phys. B135 (1978) 149.
[2] E. Cremmer, B. Julia and J. Scherk, Phys. Lett. B76 (1978) 409.
[3] M.J. Duff, J.X. Lu, Nucl. Phys. B347 (1990) 394.

 

[arivero]

What is that has to do with my post? I was making a group theoretical justification for the ''why $D =11$ in M-theory'', and the fact that $D_{min} = 11$ M-theory has $D_{max}=11, \ N =1$ supergravity as its low energy limit. What is that has to do with F-theory? Do you even know the metric signature in F-theory?.

Well, your argument was really an extra argument made by Witten more than ten years before of the discovery of M-theory and in the line of your argument is it fact very interesting to ask what happens with 8 or 9 extra dimensions instead of 7. Note that obviously the GUT group SO(10) is the group of isomorfisms of the 9-sphere and that two very popular subgroups, SU(4)xSU(2)xSU(2) and SU(5), also used in GUT theories, are groups of isomorfisms of 8-dimensional spaces, S5xS3 or CP4 respectively. In fact the 7-dimensional manifolds you mention are generically produced, in Witten's paper, quotienting S5xS3 by some U(1) symmetry. So when you are arguing in the basis of Kaluza Klein it is very natural to ask for D=12 and D=13 and wonder if the breaking or not existence of GUT groups (no proton decay) is related to the fact that sugra does not go beyond D=11.

 

[samalkhaiat]

Well, your argument was really an extra argument made by Witten

Yes, in that paper [1], Witten looked for “a realistic KK theory”, i.e., one with ground state solution given by the product $M^{4} \times B^{n}$, where $B^{n}$ is a compact n-dimensional space with symmetry group $G$. Then by demanding that $SU(3) \times SU(2) \times U(1) \subset G$, he proved that the minimum value of $n$ must be 7. Then he listed all possible 7-dimensional manifolds with $SU(3) \times SU(2) \times U(1)$ symmetry.

more than ten years before of the discovery of M-theory

Yes, this is because M-theory is not just $D=11$ KK theory. Indeed, Witten [2] showed that it is impossible to derive (by the usual KK method of compactifying on a manifold) a $D=4$ chiral theory starting from a non-chiral theory such as $D=11$ supergravity. Exactly ten years later, Horava & Witten [3] solved this problem by compactifying on something that is not a manifold. Also see [4,5].

and in the line of your argument is it fact very interesting to ask what happens with 8 or 9 extra dimensions instead of 7. Note that obviously the GUT group SO(10) is the group of isomorfisms of the 9-sphere and that two very popular subgroups, SU(4)xSU(2)xSU(2) and SU(5), also used in GUT theories, are groups of isomorfisms of 8-dimensional spaces, S5xS3 or CP4 respectively. In fact the 7-dimensional manifolds you mention are generically produced, in Witten's paper, quotienting S5xS3 by some U(1) symmetry. So when you are arguing in the basis of Kaluza Klein it is very natural to ask for D=12 and D=13 and wonder if the breaking or not existence of GUT groups (no proton decay) is related to the fact that sugra does not go beyond D=11.

There is no problem in formulating a KK type theory in any number of extra dimensions if you can get your quantum numbers correct. But you can never go beyond $D =11$ in supergravity. What would you do with the infinite tower of massless, spin $J = 2 + k$ particles and their superpartners?
You can easily see the problem in the non-linear sigma model action: An action for a membrane can be formulated in any number of spacetime dimensions. However, the supersymmetric version of the action is characterized by a local fermionic symmetry. This symmetry requires a Wess-Zumino-Witten term in the action. Such term is only possible for $D = 4, 5, 7$ and $11$ dimensional space-time [6]

[1] E. Witten (1981) “Search for a realistic Kaluza-Klein theory” Nucl. Phys. B186, 412.
[2] E. Witten (1985) “Fermion quantum numbers in Kaluza-Klein theory”: In “Quantum Field Theory and the Fundamental Problems of Physics” Ed. R. Jackiw, N. Khuri, S. Weinberg & E. Witten, MIT Press, p227.
[3] P. Horava, E. Witten (1996) Nucl. Phys. B460, 506.
[4] E. Witten (1995) “String dynamics in various dimensions” B443, 85.
[5] E. Witten (1997) “Solutions of 4-dimensional field theories via M-theory” Nucl. Phys. B500, 3.
[6] M. Henneaux & L. Mezincescu (1986) Phys. Lett. 180B, 370.