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Let $\{P_i\}$ be a subset of $SU(n)$ such that for any $U$ in another subset (or perhaps subgroup) $H$ of $SU(n)$: $$P_1UP_2U\cdots P_mU=I$$ where $I$ is the identity element. Is there a sequence $\{P_i\}$ such that $H$ be enlarged to the whole $SU(n)$? Also (perhaps more interesting): how large can $H$ be made?

Perhaps a similar question can be asked about groups in general. For $SU(n)$, I suspect that $U$'s must be generated by a single Lie Algebra element [which I understand to mean $U=exp(-i \theta A)$ for a fixed Hermitian $A$ and any real number $\theta$]. Notice that a similar question can be asked about rotation matrices in $\mathbb{R}^3$ which I believe to be sufficient for a more general proof. A direction that I pursued was to note that the transformation of the generator (logarithm) of $U$ under the sequence has to be a linear map (identically zero) and then try to prove that it (or its restriction) can have a dimension of at most 1.

As a non-$SU(n)$ example consider the group of affine transformations of the 3-dimensional Euclidean space and take ${P_i}$ to be 4 alternating 180 rotations (flips) along $x$ and $y$. Then $U$ can be any translation. Now It is highly unlikely that the set of $U$'s can be extended to the whole affine group by using a larger set of $P_i$'s but I can't prove it.

Excuse my quantum physicist notation and I hope that this is not too elementary although I would appreciate an elementary answer.

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    $\begingroup$ Consider the case $n=2$. If $P_1UP_2U=I$ held for all $U$ then ($U=P_1^{-1}$) we'd have that $P_1=P_2$ and hence $(P_1U)^2=I$, i.e. every matrix would have to have square $I$. $\endgroup$ – Tilman Jul 29 '14 at 6:56
  • $\begingroup$ @Tilman That's why you will need more than 2 of those. Also notice (question edited to emphasize this) that I am really looking for $P_i$ with such properties for the largest subgroup possible. With just 2 you can think about $P_1 = P_2$ being a 180 degree rotation around $x$ while $U$ is any rotation around $z$. In this case $H$ would be a subgroup but obviously not the whole group. $\endgroup$ – Kaveh Khodjasteh Jul 29 '14 at 14:19
  • $\begingroup$ If I'm understanding your comment and your third paragraph, you intend that the subscript $n$ in the first equation is a different integer than the $n$ in $SU(n)$. If so, it might be good to change the notation a bit. $\endgroup$ – j.c. Jul 29 '14 at 17:32
  • $\begingroup$ @j.c. thanks for the correction. I fixed it. $\endgroup$ – Kaveh Khodjasteh Jul 29 '14 at 17:52
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I can show that it is impossible to achieve $H=SU(n)$. More generally, if $G$ is a compact connected Lie group, I will show that the map $U \mapsto P_1 U P_2 U \cdots P_m U$ is surjective, and therefore the preimage of the identity can't be all of $G$.

Proof Such a $G$ is a compact connected orientable manifold so we can talk about the degree of self-maps of $G$. If $\phi: G \to G$ is a continuous function which misses a point, then $\phi$ has degree $0$. We will compute the degree of $U \mapsto P_1 U P_2 U \cdots P_m U$ and see that it is nonzero.

Since $G$ is connected, $U \mapsto P_1 U P_2 U \cdots P_m U$ is homotopic to $U \mapsto U^m$. A generic matrix in $SU(n)$ has $m^{n-1}$ many $m$-th roots in $SU(n)$ (and $m^n$ many $m$-th roots in $U(n)$). At some point in my life, I worked out that all of these $m$-th roots contribute to the degree with the same sign. More generally, let $G$ be a compact connected Lie group of rank $r$ (meaning that maximal torii have degree $r$). For a generic element $x$ of $G$, there is a unique maximal torus $T$ containing $x$ and all $m$-th roots of $x$ are in that torus; there are $m^r$ of them. For our purpose, the only important fact is that $m^r \neq 0$, so $U \mapsto U^m$ has nonzero degree and $U \mapsto P_1 U P_2 U \cdots P_m U$ must be surjective. $\square$

ADDED Robert Bryant proves that, if $G$ is a compact connected Lie subgroup and $T$ is a connected subgroup so that $P_1 U P_2 U \cdots P_m U=1$ for $U \in T$, then $T$ is contained in a maximal torus. The point of this edit is to point out that, if $G$ is semisimple, and $T$ is any maximal torus, then there are $P_i$ which achieve this. Robert Bryant already does this for $SU(n)$ in his answer.

Let $N(T)$ be the normalizer of $T$, so we have a short exact sequence $1 \to T \to N(T) \to W \to 1$, where $W$ is the Weyl group. Let $c$ be any element of the Weyl group which acts with no eigenvalues of $1$ on the Lie algebra of $T$ (for example, a Coxeter element) and let $h$ be the order of $c$. Let $P_1$ be a lift of $c$ to $N(T)$. Let $P_1^h = z$ (an element of $T$). Set $P_1=P_2 = \cdots = P_{h-1}$ and $P_h = P_1 z^{-1}$. Then I compute $\def\Ad{\mathrm{Ad}}$ $$P_1 \exp(t) P_1 \exp(t) \cdots P_1 \exp(t) P_1 z^{-1} \exp(t) = \exp\left( (\Ad(P_1) + \Ad(P_1)^1 + \cdots + \Ad(P_1)^h ) t \right) P_1^h z^{-1} = \exp(0) \cdot 1.$$

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  • $\begingroup$ That's nice. I have a feeling that I can extend this proof (basically the homotopy argument) to the following extension: $P_1e^{i t_1 h}\cdots P_me^{it_m h}=I$ for $h$ from a subset of $su(n)$, then no choice of $P_i$ and $t_i$ would eliminate result in identity regardless of $h$. $\endgroup$ – Kaveh Khodjasteh Jul 31 '14 at 1:33
  • $\begingroup$ Btw. compactness is only being used for the notion of degree, right? Otherwise it seems that one only needs $U\mapsto U^m$ to be surjective. $\endgroup$ – Kaveh Khodjasteh Jul 31 '14 at 15:16
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New answer: I now have an answer for the subgroup case that the OP originally asked about. In fact, one has the following result: Let $G$ be a connected compact Lie group and let $p = (p_1,\ldots,p_m)$ be an $m$-tuple of elements of $G$. Define the set $$ H(p) = \{ g\in G\ |\ (p_1g)(p_2g)\cdots(p_mg) = 1\}. $$ If $H\subset H(p)$ is a connected subgroup of $G$, then $H$ is abelian. In particular, $H$ lies in a maximal torus in $G$.

I am going to give the proof in Part III below. I am leaving Parts I and II (which were part of my previous answer) in place because they contribute to the answer below in some way.

Part I: In the case $G=\mathrm{SU}(n)$, one can have an $H\subset H(p)$ be a maximal torus for appropriate choice of $p$, so the abelian condition is the best possible.

Just do this: Let $T^{n-1}\subset \mathrm{SU}(n)$ be the maximal torus that consists of diagonal elements, i.e., the elements in $\mathrm{SU}(n)$ that preserve the lines $L_i=\mathbb{C}e_i\subset\mathbb{C}^n$, where $e_i$ is the standard basis of $\mathbb{C}^n$. Let $Q\in\mathrm{SU}(n)$ be the matrix that satisfies $Qe_i=e_{i+1}$ for $1\le i<n$ and $Qe_n=(-1)^{n-1}e_1$. (Note that $Q^n=(-1)^{n-1}I$.) Then, for $U\in T^{n-1}$, one has the identity $$ U\cdot {Q}UQ^{-1}\cdot {Q}^{2}UQ^{-2}\cdot\ \cdots\ \cdot {Q}^{(n-1)}UQ^{-(n-1)} = I. $$ (Each of the conjugates in the product is a diagonal matrix with the eigenvalues of $U$ cyclically permuted, and the product of the eigenvalues of $U$ is $1$.) Conjugating this identity by $Q^{n-1}$, one obtains $$ Q^{1-n}U(QU)^{n-1} = I $$ for all $U\in T^{n-1}$, which is an identity of the form the OP desired (with $m=n$).

Part II: Here is an algebraic proof that one can't have $H=\mathrm{SU}(n)=G$: Suppose that one had $$ P_1UP_2U\cdots P_mU = I $$ for all $U\in \mathrm{SU}(n)$. Then, complexifying this relation, one would have $$ P_1XP_2X\cdots P_mX = I $$ for all $X\in \mathrm{SL}(n,\mathbb{C})$. [Put another way: The mapping $f:\mathrm{SL}(n,\mathbb{C})\to \mathrm{SL}(n,\mathbb{C})$ defined by $f(X)=P_1XP_2X\cdots P_mX$ is holomorphic, but, by hypothesis, it is constant on the totally real submanifold $\mathrm{SU}(n)\subset \mathrm{SL}(n,\mathbb{C})$, which is the fixed point set of the antiholomorphic involution $X\mapsto {}^t\bar X^{-1}$. Thus, $f$ is constant.]

Multiplying by $\lambda^m\in\mathbb{C}^\ast$, one finds $$ P_1(\lambda X)P_2(\lambda X)\cdots P_m(\lambda X) = \lambda^m\ I $$ for all $X\in \mathrm{SL}(n,\mathbb{C})$ and $\lambda\in\mathbb{C}^\ast$, which implies that $$ P_1ZP_2Z\cdots P_mZ = \det(Z)^{m/n}\ I $$ for all $n$-by-$n$ complex matrices $Z$ with nonzero determinant. Since the left hand side is a polynomial in the entries of $Z$, it follows that $m/n = k$ for some integer $k$, so $$ P_1ZP_2Z\cdots P_mZ = \det(Z)^k\ I $$ for all $n$-by-$n$ complex matrices $Z$, where $m=kn$. Now, choose $Z$ to be rank $1$, say, $Z=x\ {}^ty$ for $x,y\in \mathbb{C}^n$. Taking the trace of the above relation and using the fact that $\det(Z)=0$, one finds that the following product of quadratic forms must vanish identically $$ ({}^tyP_1x)\ ({}^tyP_2x)\ ({}^tyP_3x)\ \cdots ({}^tyP_mx) = 0. $$ Since $x$ and $y$ are arbitrary, it follows that one of the factors ${}^tyP_jx$ must vanish identically. But this is clearly impossible, since each $P_j$ is invertible.

Remark: In fact, this proof works for any of the real forms of $\mathrm{SL}(n,\mathbb{C})$, such as $\mathrm{SL}(n,\mathbb{R})$, $\mathrm{SU}(k,n{-}k)$, or $\mathrm{SU}^\ast(n/2) = \mathrm{SL}(n/2,\mathbb{H})$, since they all have the same complexification.

Part III: Now, finally, the general proof that $H$ must be abelian.

First, by embedding $G$ into some $\mathrm{SU}(n)$ for some $n$ sufficiently large, I can assume that $G=\mathrm{SU}(n)$. Next, if $p=(p_1,\ldots,p_m)$ were such that $H(p)$ contained a connected nonabelian subgroup $H\subset G$, then, by replacing $H$ by its closure, I can assume that $H$ is a connected compact nonabelian Lie subgroup of $G$. In particular, it follows from the classification of compact Lie groups that $H$ contains a Lie subgroup $K$ that is isomorphic to either $\mathrm{SO}(3)$ or to $\mathrm{SU}(2)$.

Now recall that every compact simple Lie group $G$ has a closed, left-invariant Cartan $3$-form $\gamma_G$, whose value at the identity is $\gamma_G(x,y,z) = \kappa\bigl([x,y],z\bigr)$ for $x,y,z\in {\frak{g}}=T_eG$, where $\kappa$ is the Killing form of $\frak{g}$. This form generates the deRham cohomology group $H^3_{dR}(G)\simeq \mathbb{R}$. When $K=\mathrm{SO}(3)$ or $\mathrm{SU}(2)$, the Cartan $3$-form $\gamma_K$ is a canonical volume form, and, whenever $h:K\to G$ is a nonconstant homomorphism, one has $h^*(\gamma_G) = n_h\ \gamma_K$ for some integer $n_h>0$.

Now, consider the mapping $\mu:K\to G=\mathrm{SU}(n)$ defined by $$ \mu(k) = (p_1k)(p_2k)\cdots(p_mk). $$ By hypothesis, $\mu(k) = 1$ for all $k\in K$. However, it is straightforward to compute that $$ \mu^*\bigl([\gamma_G]\bigr) = m\ n_h\ [\gamma_K]\not = 0, $$ so it is not possible for $\mu$ to be constant.

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  • $\begingroup$ Nice alternate proof! But is complexification more elementary than degree theory? By the completely arbitrary standard of what year I learned them, I claim that degree theory is 2 years simpler :). $\endgroup$ – David E Speyer Jul 30 '14 at 20:03
  • $\begingroup$ @DavidSpeyer: I think it's more a matter of taste than 'elementary' versus 'advanced'. 'Complexification' in this instance is straightforward because $\mathrm{SU}(n)$ is already embedded as a totally real submanifold of $\mathrm{SL}(n,\mathbb{C})$, in fact, as the fixed point set of an anti-holomorphic involution. OTOH, verifying that the $m$-th roots of a generic point contribute the same sign to the degree does require a computation (as you hinted in your answer), and I think you still have to do something to verify this. (Doesn't degree theory require homology to prove it is well defined?) $\endgroup$ – Robert Bryant Jul 30 '14 at 21:53
  • $\begingroup$ I completely agree it is a matter of taste. The way I learned it was morally de Rham cohomology, but proved without developing the whole theory: Given a map $f: X \to Y$ and two points $y_1$ and $y_2$ in $Y$, find a path $\gamma$ connecting the $y_i$. Write down an $n-1$ form $\eta$ in a neighborhood of $\gamma$ so that $d \eta$ is concentrated on small neighborhoods of $y_1$ and $y_2$. Then $\int_X f^{\ast} (d \eta)$ must be $0$. Shrinking $\eta$ closer and closer to $\gamma$ shows that $\# f^{-1}(y_1)=\# f^{-1}(y_2)$ (counted with multiplicity). $\endgroup$ – David E Speyer Jul 30 '14 at 22:00
  • $\begingroup$ I marked David Speyers answer as correct but clearly your algebraic alternative proof is also great. To me they both required me the same amount of time to parse (a long time that is) but I never took topology or linear algebra ;) $\endgroup$ – Kaveh Khodjasteh Jul 31 '14 at 13:59
  • $\begingroup$ @KavehKhodjasteh: They both have their advantages. Actually, this morning (about an hour ago), I posted a more-or-less complete answer to your question of which connected subgroups $H$ of $\mathrm{SU}(n)$ could occur (and even for general compact connected Lie groups rather than just for $\mathrm{SU}(n)$); the answer is that they have to be abelian, and hence lie in a maximal torus. However, it has to clear the moderators before it allowed to appear. (I'm afraid that my obsessive editing to fix small typos, etc. has triggered some protective feature I wasn't aware of until now.) $\endgroup$ – Robert Bryant Jul 31 '14 at 14:18

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