# Real varieties with enough algebraic loops

Let $(X,\sigma)$ be a complex variety with complex conjugation (equivalently, an algebraic variety over $\mathbb R$). We use the notations $X(\mathbb R):=X^\sigma$ for the set of fixed points of $X$ under $\sigma$. Equivalently, this is the set of real points of the algebraic variety $X$.

Here's a property which I believe compact semisimple real algebraic groups to have (but not $S^1$!):

Any algebraic map from $S^1:=(\mathbb C^\times,z\mapsto \bar z^{-1})$ to $(X,\sigma)$ induces a $\mathit C^\infty$ map from $S^1(\mathbb{R})$ to $X(\mathbb{R})$. Let us say that $X$ has enough algebraic loops if these algebraically induced are dense in the space of all $\mathit C^\infty$ maps from $S^1(\mathbb{R})$ to $X(\mathbb{R})$, equipped with the $C^\infty$ topology.

• Is it true that $(G,\sigma)$ has enough algebraic loops? Here, $G$ is a complex semisimple algebraic group and $\sigma$ its Cartan involution (so that $G^\sigma$ is the compact form of $G$)

• Note that the spheres $S^n$ (see David's comment for their precise definition) seem to have enough algebraic loops iff $n\not = 1$. Indeed, this would follow from the previous statement given that they are homogeneous spaces. Is this indeed true?

• Is there a reasonably large class of real algebraic varieties for which one can prove that they have enough algebraic loops?

• I'm confused when you say "Let $(X,\sigma)$ be a complex variety with complex conjugation (equivalently, a variety over $\mathbb{R})$". The sphere $S^n$ is a real variety but it is not almost-complex, unless $n=2$ or $n=6$. I'm missing something about this "equivalently"? – Francesco Polizzi Feb 20 '14 at 7:34
• The complex variety in question is $\{ (z_0, z_1, \ldots, z_n) : z_0^2+z_1^2+\cdots + z_n^2=1 \}$, the involution $\sigma$ is $(z_0, z_1, \ldots, z_n) \mapsto (\overline{z}_0, \overline{z}_1, \ldots, \overline{z}_n)$. – David E Speyer Feb 20 '14 at 10:43
• Just to check, $S^1$ in this context means $\{ (x,y) \in \mathbb{R}^2 : x^2+y^2=1 \}$, right? And your maps must be given by polynomials in $x$ and $y$, not just rational functions whose poles are disjoint from the real locus. For example, $(x,y) \mapsto 1/(2-x^2-y^2)$ is not okay, right? – David E Speyer Feb 20 '14 at 15:13
• @Jason: I am aware of the papers that prove that the inclusion of algebraic loop groups into smooth loop groups are homotopy equivalences. This is done by computing cohomology on both sides (by different methods) and checking that the induced map is an isomorphism. Density is a much stronger requirement, and I'm wondering if it also holds. @ David and Jason: you are both right! I tried to edit my question to make it less confusing. I hope that I didn't make things any worse (feel free to edit). – André Henriques Feb 20 '14 at 19:49
• @AndréHenriques: Frankly I don't see how to get density for, say, the compact-open topology. However, for every finite subset of $\mathbf{S}^1$ (unfortunately not all compact subsets), perhaps we can prove existence of a (real) algebraic morphism with specified values on that finite set. This is related to "strong approximation" of adelic points by integral points (for the "integer ring" $\mathbb{R}[z,z^{-1}]$). Much is known about strong approximation for algebraic groups. – Jason Starr Feb 20 '14 at 23:55

Obviously, if $X$ is complete, you can just speak about maps $\mathbb{P}^1\to X$, or you can also replace $\mathbb{P}^1$ with the "standard" circle (conic in $\mathbb{P}^2$). This approximation problem used to be popular, see e.g. the survey

MR2882786
Bochnak, Jacek; Kucharz, Wojciech
Algebraic approximation of smooth maps.
Univ. Iagel. Acta Math. No. 48 (2010), 9–40.

Right away, in an earlier paper

MR0873026
Bochnak, J.; Kucharz, Wojciech
Algebraic approximation of mappings into spheres.
Michigan Math. J. 34 (1987), no. 1, 119–125

they give an affirmative answer to your question (among others) for $X=S^n$ with $n=1,2,4$. (Accidentally, the survey claims that their conjecture is still open for other values of $n$. Though, the conjecture is more general that your question about $S^1$.)

• The second paper cited is studying "entire rational maps" not polynomial maps. In the context of our problem, write $S^1 = \{ (x,y) \in \mathbb{R}^2 : x^2+y^2=1 \}$. The components of their maps $S^1 \to Y$ will be rational functions $f(x,y)/g(x,y)$ so that $g$ is nowhere vanishing on $S^1$. Andre wants honest polynomials. – David E Speyer Feb 20 '14 at 15:09
• The same comment applies to the first paper. – David E Speyer Feb 20 '14 at 15:22
• @DavidSpeyer: I do not believe that Henriques wants "honest polynomials". I believe he wants morphisms, $f:\mathbb{C}^\times \to X$ such that $\sigma \circ f$ equals $f\circ \tau$, where $\tau(z) = \overline{z}^{-1}$. So $f$, considered as a rational map from $\mathbb{P}^1$ to some completion of $X$, is allowed to have arbitrary poles at the (conjugate) pair $\{0,\infty\}$. Of course, for $X$ that are already complete, as in Degtyarev's answer, there are no poles at all (because $X$ is complete). That is why these "entire rational maps" are "entire". – Jason Starr Feb 20 '14 at 16:18
• I agree that this distinction would not be interesting if $X$ were proper, but Andre's proposed examples are algebraic groups whose $\mathbb{R}$-points are compact, and hence affine varieties. For example, Andre writes that an arbitrary map $S^1 \to S^1$ cannot be approximated by algebraic maps. This is true with my interpretation (the only algebraic maps are $z \mapsto z^n$ for $n$ an integer) but it is false if we allow maps with other complex poles. – David E Speyer Feb 20 '14 at 18:21
• Let me confirm: David Speyer is right. I care about morphisms defined on all of $\mathbb C^\times$. So the only poles should be at $0$ and at $\infty$. – André Henriques Feb 20 '14 at 19:42

Here is a sketch of a possible proof $L^{\infty}$ (uniform convergence) approximation of continuous maps to $S^3 = SU(2)$. (Actually, I need my maps to be a little better than continuous, because I need them not to be space filling curves.) As Andre observes, this also implies the result for $S^2$.

I suspect that someone with a stronger background in topology could deal with the space filling curves. I can't decide whether or not I think we should be able to push $L^{\infty}$ to $C^{\infty}$.

We will describe $S^3$ as $\{ (w,x,y,z) : w^2+x^2+y^2+z^2 = 1 \}$. We'll also think of this as the unit quaternions, writing $w+xi+yj+zk$. Stripped of the sophisticated language, the question is the following: We have a smooth map $\gamma: S^1 \to S^3$. Write $\gamma(\theta) = (w(\theta), x(\theta), y(\theta), z(\theta))$. Can we approximate $\gamma$ by maps $\gamma_1(\theta) = (w_1(\theta), x_1(\theta), y_1(\theta), z_1(\theta))$ where $w_1$, $x_1$, $y_1$ and $z_1$ are given by Fourier series of finite length, and the identity $$w_1^2 + x_1^2 + y_1^2 + z_1^2 = 1$$ must hold identically.

Since $\gamma$ is not a space filling curve, it misses some point of $S^3$ and we will assume without loss of generality that it misses the quaternion $-1$. We can define the logarithm on $SU(2) \setminus \{ -1 \}$: The logarithm of $w+xi+yj+zk$ is the unique quaternion $pi+qj+rk$ with $\exp(pi+qj+rk) = w+xi+yj+zk$ and $p^2+q^2+r^2 < \pi^2$. For $a \in \mathbb{R}$, and $w+xi+yz+zk \neq -1$, we will define $(w+xi+yz+zk)^a$ to be $\exp(a \log(w+xi+yz+zk))$.

Choose a large integer $n$. For $0 \leq s \leq n-1$, define $a_s : S_1 \to \mathbb{R}$ to be the piecewise linear function which is $1/2$ on $[2 \pi (s-1/2)/n, 2 \pi (s+1/2)/n]$, interpolates linearly between $0$ and $1/2$ on $[2 \pi (s-3/2)/n, 2 \pi (s-1/2)/n]$ and $[2 \pi (s+1/2)/n, 2 \pi (s+3/2)/n]$m and is $0$ on the rest of $S^1$. Define $g_s(\theta) = \gamma(2 \pi s/n)^{a_s(\theta)}$ and define $g(\theta) = \prod_{s=1}^n g_s(\theta)$. As $n \to \infty$, the function $g(\theta)$ uniformly approaches $\gamma(\theta)$. Therefore, it is enough to choose $n$ so that $g$ is within $\epsilon/2$ of $\gamma$, and then approximate each $g_s$ within $\epsilon/(2n)$ by finite Fourier series.

If we wanted to do $C^{\infty}$ approximation rather than $L^{\infty}$, we would of course want the bump functions $a_s$ to be smooth rather than $C^{\infty}$. I expect this isn't hard, but it's not what I want to focus on.

In short, we have reduced to approximating maps of the form $\theta \mapsto q^{a(\theta)}$ where $q$ is a fixed quaternion not equal to $-1$. We may conjugate $q$ to be of the form $a+bi$, so we need to approximate $w(\theta) + x(\theta) i$ by $w_1(\theta) + x_1(\theta) i + y_1(\theta) j + z_1(\theta) k$.

Take $w_1$ and $x_1$ to be finite Fourier series approximations of $w$ and $x$, say within $\epsilon^2/100$. Multiplying by a scalar near $1$, we can further arrange that $w_1^2+x_1^2 \in (1-\epsilon^2/20, 1-\epsilon^2/10)$. Define $u(\theta) = 1-w_1(\theta)^2-x_1(\theta)^2$.

We are now reduced to finding finite Fourier series $y_1(\theta)$ and $z_1(\theta)$ with $y_1^2 + z_1^2 = u$. We no longer need to impose that $y_1$ and $z_1$ are close to $y=z=0$, since this is automatic from the identity $y_1^2 + z_1^2 = u < \epsilon^2/10$.

Lemma Let $u: S^1 \to \mathbb{R}_{>0}$ be a finite Fourier series. Then there are finite Fourier series $y_1$ and $z_1$ with $u=y_1^2+z_1^2$.

Proof: We will write $u$ as $\sum u_m q^m$, where $q=e^{i \theta}$. The condition that $u$ is real valued means that $u_{-m}=\overline{u_m}$.

Considering $u(q)$ as a function of a complex variable $q$, we have $u(\overline{q}^{-1}) = \overline{u(q)}$. So, if $\alpha$ is a zero of $u$, so is $\overline{\alpha}^{-1}$. Also, none of these zeroes can be on the unit circle, since $u$ takes positive values. Therefore we can write $$u(q) = c \prod (1-q/\alpha_i )(1-q^{-1}/\overline{\alpha_i})$$ for some positive constant $c$ and some list of zeroes $\alpha_i$.

Define $v(q) = \sqrt{c} \prod (1-q/\alpha_i)$; then we have $v(e^{i \theta}) \overline{v(e^{i \theta})} = u(\theta)$. Take $y_1$ and $z_1$ to be the real and imaginary parts of $v$. $\square$

Note that this proof gives us no control over the sizes of $y'_1$ and $z'_1$, which is why I can't decide whether we should be able to get convergence in $C^k$ for $k \geq 1$.

We can also approximate maps $S^1 \to SO(3)$. Multiplying $\gamma$ by $\begin{pmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$ if necessary, we may assume that $\gamma$ has a square root $\sqrt{\gamma}: S^1 \to SU(2)$. Approximate $\sqrt{\gamma}$ by polynomials and then observe that the squaring map $SU(2) \to SO(3)$ is given by polynomials (I wrote them down here.)

I think this argument might actually be extendible to all semisimple compact Lie groups. Here is a crude sketch.

The bump function trick let's us focus on approximating maps $g$ that land in the exponential of a one dimensional subspace $\mathfrak{s}$ of the Lie algebra. Let $T$ be the closure of $\exp(\mathfrak{s})$, so $T$ is a torus; let its Lie algebra be $\mathfrak{t}$. Multiplying by a cocharacter $S^1 \to T$, we may assume that $g: S^1 \to T$ is null homotopic and thus has a logarithm $\log g: S^1 \to \mathfrak{t}$. Let $\mathfrak{r}_1$, $\mathfrak{r}_2$, ...., $\mathfrak{r}_m$ in $\mathfrak{t}$ be the one dimensional subspaces spanned by the simple roots: we can write $\log g$ as a linear combination $\log g = \sum h_i$ of maps $h_i: S^1 \to \mathfrak{r}_i$ and can exponetiate these to maps $\exp(h_i)$ from $S^1$ to one dimensional torii $R_i$. Each $R_i$ sits in an $SU(2)$ or $SO(3)$, so we can approximate $\exp(h_i)$ by the above argument.

• This is a beautiful argument! Your Lemma is particularly pretty, and it's really a shame that it allows for wild oscillations in the directions of the $y_1$ and $z_1$ coordinates. I also completely agree with your last paragraph that claims that this argument easily extends to the case of arbitrary semisimple groups. – André Henriques Feb 22 '14 at 9:29
• Thanks! The lemma is a modification of a standard proof that a positive polynomial is a sum of squares of two polynomials. Once $u$ is fixed by all the prior choices, there are only finitely many options for $y_1$ and $z_1$, so it makes sense that we can't control their oscillations. But there are so many arbitrary choices before that point, that it certainly seems like we could maneuver to end up with a good $u$ somehow! – David E Speyer Feb 22 '14 at 14:00
• In the proof that $\mathrm{SO}_3$ has enough loops, I do not understand why the continuous $\gamma\colon S^1\rightarrow \mathrm{SO}_3$ has a lift $\sqrt\gamma$ to $\mathrm{SU}_2$. After all, $\mathrm{SO}_3$ is homeomorphic to $\mathbf P^3(\mathbf R)$, and $\gamma$ can be homotopically nonzero. Why does it lift to the universal covering? – Johannes Huisman Jan 28 '16 at 19:50

Here is an argument that $S^3=SU(2)$ has enough algebraic loops.

First of all, it is enough to consider loops that don't hit the element $-1\in SU(2)$ since they form a dense subset in the space of all loops.

Given a $C^\infty$ loop $\gamma:S^1\to SU(2)$, consider its preimage $\tilde\gamma:S^1\to \mathfrak{su}(2)$ along the exponential map $exp:\mathfrak{su}(2)\to SU(2)$. By standard Fourier analysis, there exist good approximations of $\tilde\gamma$ by trigonometric polynomials.

Now here's an observation: $$(\cos(n\theta),0,0)={\textstyle\frac12} [(\cos(n\theta),\sin(n \theta),0) + (\cos(-n\theta),\sin(-n \theta),0)]$$ and similarly for $(\sin(n\theta),0,0)$. Therefore, we may approximate $\tilde \gamma$ by a finite linear combination $$\tilde\gamma\,\, \approx\,\, \sum_{i=1}^n f_i$$ of functions $f_i:S^1\to \mathfrak{su}(2)=\mathbb R^3$ that are of the form $$f_i(\theta)=a_i(\pm\cos(n_i\theta),\pm\sin(n_i \theta),0)$$ up to permutation of the three coordinates of $\mathbb R^3$.

Now the crucial observation is that $$exp(f_i):S^1\to SU(2)$$ is algebraic.

To finish the argument, pick $N\gg 1$, and consider the product $$\textstyle\big(exp(\frac{f_1}N)exp(\frac{f_2}N)\ldots exp(\frac{f_n}N)\big)^N$$ in the group. For big $N$, that expression is an excellent approximation of $\gamma=exp(\tilde \gamma)$.

To deal with the case of arbitrary compact connected Lie group $G$, one can argue similarly to the way David did in his answer: the smooth loop group of $G$ contains many copies of $LSU(2)$ and $LSO(3)$ in it, and is generated by them.
The case of homogeneous spaces is then an easy consequence.

• This is really nice! I'm glad I finally noticed it. – David E Speyer Jul 30 '15 at 14:19