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There is a nice smooth retraction from $\operatorname{GL}(n,\mathbb{C})$ onto $\operatorname{U}(n)$, which can be explained using polar decomposition. There is an analogous one from $\operatorname{GL}(n, \mathbb{R})$ onto $\operatorname{O}(n)$. Both are "natural" from a Lie theoretic point of view (in a sense that shall not be made precise for now).

I am wondering what is the largest subgroup of $\operatorname{GL}^{+}(7,\mathbb{R})$ ('+' here just means with positive determinant) which smoothly retracts onto $\operatorname{G}_2$?

I have a strange feeling it could be $\operatorname{G}_2$ itself, because I don't think that one can split a $G_2$ structure into 2 independent pieces, one being a Riemannian inner product, and something else. For instance if one uses the usual $3$-form as one of the pieces, then the subgroup of $\operatorname{GL}^{+}(7,\mathbb{R})$ which preserves that $3$-form is $G_2$ itself (so that preserving that $3$-form implies preserving an inner product).

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  • $\begingroup$ You really mean "retract" and not "retract by deformation"? I understand "smoothly retract" as the existence of a smooth map from the group $H$ (containing $G_2$) to $G_2$ that is the identity on $G_2$. By smooth retract by deformation, I would understand a map $f:H\times [0,1]\to H$ such that $f(g,t)=g$ for all $t$ and all $g\in G_2$, $f(-,0)$ is the identity and $f(-,1)$ maps into $G_2$. $\endgroup$
    – YCor
    Commented Oct 9, 2016 at 20:45
  • $\begingroup$ I just want a smooth retraction, not deformation retraction, which has nice equivariance properties, if you will. $\endgroup$
    – Malkoun
    Commented Oct 9, 2016 at 20:48
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    $\begingroup$ I guess $G_2$ stands for the compact form endowed with its standard 7-dim representation, right? Surely you can take the group generated by it and the (positive) scalar matrices. I guess that's it. Our mysterious group cannot have a unipotent radical (the $G_2$-rep is irreducible) and a reductive group is homotopically equivalent to its maximal compact subgroup. $\endgroup$
    – Uri Bader
    Commented Oct 9, 2016 at 20:50
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    $\begingroup$ OK. The list of subgroups of $SL_7$ containing $G_2$ is very short. The only compact ones are $G_2$ itself and $SO(7)$. The non-compact ones are only $SL_7$ itself (unless the complex $G_2$ occurs as a 28-dimensional real subgroup here? I doubt this but this would not matter). So the question is only whether $SO(7)$ retracts. $\endgroup$
    – YCor
    Commented Oct 9, 2016 at 20:58
  • $\begingroup$ @Uri Bader, I think you are right about the mysterious subgroup. $\endgroup$
    – Malkoun
    Commented Oct 9, 2016 at 21:19

1 Answer 1

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There is no retraction of $\mathrm{SO}(7)$ onto $\mathrm{G}_2$. If such a retraction $\rho:\mathrm{SO}(7) \to \mathrm{G}_2$ existed, then the composition $$ \mathrm{G}_2 \hookrightarrow \mathrm{SO}(7)\ {\buildrel{\rho}\over{\rightarrow}} \ \mathrm{G}_2\,, $$ would induce a composition of the homotopy group homomorphisms $$ \pi_6\bigl(\mathrm{G}_2\bigr) \rightarrow \pi_6\bigl(\mathrm{SO}(7)\bigr)\ {\buildrel{\rho_*}\over{\rightarrow}} \ \pi_6\bigl(\mathrm{G}_2\bigr) $$ that was the identity. However, according to M. Mimura, The homotopy groups of Lie groups of low rank, J. Math. Kyoto Univ. 6 (1967), 131–176 (see the tables on page 132), $$ \pi_6\bigl(\mathrm{G}_2\bigr)\simeq \mathbb{Z}_3 \quad\text{while}\quad \pi_6\bigl(\mathrm{SO}(7)\bigr)\simeq 0\,, $$ so this is impossible.

Beyond this, though, I feel that the OP's given motivation for asking the question suggests that the issue of concern is not really whether $\mathrm{SO}(7)$ retracts onto $\mathrm{G}_2$.

The usual reason that one wants to retract $\mathrm{GL}(n,\mathbb{C})$ onto $\mathrm{U}(n)$ and $\mathrm{GL}(n,\mathbb{R})$ onto $\mathrm{O}(n)$ is that these are the fundamental steps in the proof that a complex vector bundle over a paracompact base can always be given a unitary structure and a real vector bundle over a paracompact base can always be given an Euclidean structure (i.e., a positive definite inner product field).

The analogous question concerning $\mathrm{SO}(7)$ and its subgroup $\mathrm{G}_2$ would be whether every oriented $7$-plane bundle $E\to M$ endowed with an Euclidean structure can be given a $\mathrm{G}_2$-structure, i.e., whether there exists a 'definite' (i.e., a $\mathrm{G}_2$-defining) section of $\Lambda^3(E)\to M$. Now, the answer to this question is a definite 'no'. For example, let $M^7$ be any orientable-but-not-spinnable $7$-manifold, say $M^7 = \mathbb{CP}^2\times S^3$. Then $E=TM\to M$ certainly has a $\mathrm{SO}(7)$-structure (defined by the orientation and any Riemannian metric on $M$), but, because $M$ is not spinnable (because $\mathbb{CP}^2$ is not), its tangent bundle does not allow a structure reduction to $\mathrm{G}_2$. (This is because $\mathrm{G}_2$ is simply-connected, so that any $7$-manifold whose tangent bundle has a reduction of structure to $\mathrm{G}_2$ would necessarily be spinnable.)

Thus, this particular problem does not motivate the retraction question.

The OP's other remark expressing doubts about whether a $\mathrm{G}_2$-structure can be split into two 'independent pieces' seems to be asking for an analog of what happens in complex and symplectic geometry, where one has the intersections $$ \mathrm{U}(n) = \mathrm{GL}(n,\mathbb{C})\cap \mathrm{Sp}(n,\mathbb{R}) = \mathrm{GL}(n,\mathbb{C})\cap \mathrm{O}(n) = \mathrm{Sp}(n,\mathbb{R})\cap \mathrm{O}(n), $$ so that one can think of a unitary structure as a combination of two 'independent' (but compatible) weaker structures. The natural analog of this in the present context would be whether $\mathrm{G}_2$ can be expressed as the intersection of two (or more?) Lie subgroups of $\mathrm{GL}(7,\mathbb{R})$ that properly contain $\mathrm{G}_2$. The answer to this is a(n uninteresting) 'yes'. There are only three connected Lie groups that lie strictly between $\mathrm{G}_2$ and $\mathrm{GL}^+(7,\mathbb{R})$, and they are $$ \mathrm{R}^+{\cdot}\mathrm{G}_2,\quad \mathrm{SO}(7),\quad\text{and} \quad \mathrm{R}^+{\cdot}\mathrm{SO}(7). $$ Thus, the only nontrival way to do this is the uninteresting intersection $$ \mathrm{G}_2 = \mathrm{R}^+{\cdot}\mathrm{G}_2 \cap \mathrm{SO}(7). $$ Thus, the largest (connected) subgroup of $\mathrm{GL}^+(7,\mathbb{R})$ that retracts onto $\mathrm{G}_2$ is $\mathrm{R}^+{\cdot}\mathrm{G}_2$.

As a final argument as to why the retraction question seems to be irrelevant to the OP's apparent concerns, consider the analogous question of whether $\mathrm{SO}(2n)$ can be retracted onto $\mathrm{U}(n)$ for $n>1$. It cannot, because, such a retraction $\rho$, if it existed in the form the OP desired, i.e., $$ \mathrm{U}(n)\hookrightarrow \mathrm{SO}(2n)\ {\buildrel{\rho}\over{\rightarrow}} \ \mathrm{U}(n), $$ where the composition is the identity on $\mathrm{U}(n)$, then the composition of the induced map on fundamental groups $$ \mathbb{Z}\to \mathbb{Z}_2\to \mathbb{Z} $$ would be an isomorphism, which is manifestly impossible.

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  • $\begingroup$ Awesome answer :). Thank you so much Prof. Bryant. You read me correctly :). I was trying out a "dead alley" so to speak. I had realized that early on, though I did not have a proof that it was a "dead alley", but I had moved on, and succeeded in doing what I wanted to do. You are so smart, not just in Mathematics, but also in Psychology :). I will of course accept this brilliant answer, but please fix the small typo: you wrote $\mathbb{C}P^4$ instead of $\mathbb{C}P^2$ in the definition of $M^7$. Thank you. $\endgroup$
    – Malkoun
    Commented Dec 11, 2016 at 9:03
  • $\begingroup$ @Malkoun: You are welcome. It was an interesting question, even though I felt that it was not the 'right' one. I have fixed the typo; thanks for pointing it out. $\endgroup$
    – Robert Bryant
    Commented Dec 11, 2016 at 9:20
  • $\begingroup$ I added a link to the article (which is open access). $\endgroup$
    – David Roberts ♦
    Commented Dec 11, 2016 at 9:31
  • $\begingroup$ @Malkoun, for future reference, you can make edits yourself (subject to approval of a higher-reputation user). $\endgroup$
    – LSpice
    Commented Dec 11, 2016 at 17:32
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    $\begingroup$ @Malkoun: I think I see what you want: How's this? Let $\times:\mathbb{R}^7\times\mathbb{R^7}\to\mathbb{R}^7$ be the cross product that $G_2\subset\mathrm{SO}(7)$ preserves. Given a basis $v_i$ of $\mathbb{R^7}$ such that $v_3$ does not lie in the $3$-plane spanned by $v_1,v_2, v_1{\times}v_2$, let $e_1,e_2$ be the $GS$-orthonormalization of $(v_1,v_2)$ and write $v_3 = a\,e_1{\times}e_2 + b\,e_3$ where $b>0$ and $(e_1,e_2,e_3)$ are orthonormal and complete to an orthonormal basis by setting $e_4 = e_1{\times}e_2$, $e_5=e_2{\times}e_3$, $e_6=e_3{\times}e_1$, and $e_7=e_6{\times}e_2$. $\endgroup$
    – Robert Bryant
    Commented May 10, 2018 at 18:39

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