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Thanks!. I slightly edited the first sentence for links:
We do have Euclidean group, albeit just a stub.
And, due to the dominance of logicians around here, for better or worse our lattice points to the order-theoretic concept, while for the “common” meaning we need to type lattice in a vector space.
expanded the Idea/Definition-section, in particular added this diagram:
$\array{ & 1 && 1 \\ & \downarrow && \downarrow \\ {\text{normal subgroup} \atop \text{lattice of translations}} & N &\subset& E & {\text{translation} \atop \text{group}} \\ & \big\downarrow && \big\downarrow \\ {\text{crystallographic} \atop \text{group}} & S &\subset& Iso(E) & {\text{Euclidean} \atop \text{isometry group}} \\ & \big\downarrow && \big\downarrow \\ {\text{point} \atop \text{group}} & G &\subset& O(E) & {\text{orthogonal} \atop \text{group}} \\ & \downarrow && \downarrow \\ & 1 && 1 }$Then I added the remark (here) that the normality of the subgroup $N$ is what makes the action of the point group $G$ descend to the torus $E/N$.
Is there a good name for these tori $E/N$ equpped with actions of Euclidean point groups $G$? They want to be called “representation tori” to go along with representation spheres, but I guess nobody says that.
started to add various references.
Added a long quote from
on the history and current scope of the classification.
Not done yet, but need to interrupt now.
added pointer to
am trying to find out which of the finite subgroups of SU(2) in the D- and E-series arise, via their canonical action on $\mathbb{C}^2 \simeq_{\mathbb{R}} \mathbb{R}^4$, as point groups of crystallographic groups in 4d.
Is it just a matter of going through a list? The list in ’Summary’ tells you which are non-crystallographic.
Yes, it should be. But I have trouble parsing the entries.
Does $D_n$ denote the dihedral group? Does the binary dihedral group appear anywhere?
So I suppose I can easily check by direct inspection that the following are crystallographics groups $N \rtimes G$ in 4d, with $N = \mathbb{Z}^4$ the canonical translational subgroup of $\mathbb{R}^4$ and the point group $G$ acting on $\mathbb{R}^4\simeq_{\mathbb{R}} \mathbb{H}$ via their inclusion as subgroups of $Sp(1)$:
$N \rtimes G = ...$
$\mathbb{Z}^4 \rtimes \mathbb{Z}_2$
$\mathbb{Z}^4 \rtimes \mathbb{Z}_4$
$\mathbb{Z}^4 \rtimes Q_8$
So these must appear somewhere in these lists. But where? I must be missing something.
The term ’binary dihedral group’ shows up on the 3d page, but not given in Coxeter notation. Coxeter notation seems confusing.
Thanks for your comment. If there is an error I’ll want to fix it.
Let me see…
Hm, checking again, it looks to me like the statement in the entry agrees with the terminology associated with Bieberbach’s first theorem; fully explicitly so in the way it is stated in the recent arXiv:1907.02021, Theorem 2.3 and paragraph below (on p. 4).
Also arXiv:1208.5055 seems to agree, less authoritatively but all the more seemingly regarding it as standard: around (0.2) on p. 3.
Finally, I wouldn’t think that what you mention regarding point groups possibly containing screw-axis transformaions etc. is in contradiction to this statement. I suppose the “fractional” in your “fractional lattice vectors” signifies this.
But if you think I am missing your point, please say so and let’s try to sort this out.
Thank you for your answer!
I’m sorry, I am not sure I understand you properly. I tried reading the references you provide, and they support the definition you provide on the site, but I am afraid I am not able to wrap my head around the implications of requiring $N$ to be a free normal abelian subgroup (mostly whether “free” is a constraining hypothesis or not). But the question I am asking fortunately should not need this concept, if I got things right.
Let me rephrase my previous post to be sure we agree on the question, and that I am not making a mistake in my thinking.
Taking a non-symmorphic space group $S$ means that there is a symmetry operation $(R|a)$ that contains both an orthogonal $R$ and a translation part $t_a : x \mapsto x+a$, where the translation $t_a$ does not belong to the lattice $N$ (in a simple counter-example I saw, $a$ was half of a lattice vector, and that’s what I meant by fractional, it was not very clear, my bad).
Hence, the quotient $S/N$ would have an equivalence class corresponding to $(R|a)$, which is not an element of the orthogonal group. Is that correct?
That seems to be in contradiction with the definition of point groups I’ve read in crystallography physics books which requires them to be a subgroup of the orthogonal group, and to the inclusion that is written in the short exact sequences on the page.
It is probably a misunderstanding on my part, apologies if that is the case.
Best regards, Sami Siraj-Dine
Thanks for your reply. Now I see more clearly what you are saying.
First regarding the “free”-ness condition in the normal subgroup: That’s just part of the very definition of what a discrete translation group should be. It just means that the translation group is isomorphic to $\mathbb{Z}^n$, and that’s what we want by definition when speaking about crystals.
Regarding your worries about the quotient: While I don’t know off the top of my head if the quotient of a crystallographic group by any normal translations subgroup is always guaranteed to be a subgroup of the orthogonal groups (you seem to suspect it is not), this is also not what is being claimed in the entry, is it. In the entry we demand explicitly the horizontal inclusions shown in the diagram reproduced now:
$\array{ & 1 && 1 \\ & \downarrow && \downarrow \\ {\text{normal subgroup} \atop \text{lattice of translations}} & N &\subset& E & {\text{translation} \atop \text{group}} \\ & \big\downarrow && \big\downarrow \\ {\text{crystallographic} \atop \text{group}} & S &\subset& Iso(E) & {\text{Euclidean} \atop \text{isometry group}} \\ & \big\downarrow && \big\downarrow \\ {\text{point} \atop \text{group}} & G &\subset& O(E) & {\text{orthogonal} \atop \text{group}} \\ & \downarrow && \downarrow \\ & 1 && 1 }$Or maybe that point needs to be stated better? If that’s what it is, I’d be happy to try to improve the text in this regard.
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