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I read the Wikipedia page about root systems. It is very easy to verify that the 2 following images are showing some root systems.

https://en.wikipedia.org/wiki/Root_system

enter image description here

enter image description here

These 2 pictures are from Wikipedia. I think if we rotate one of them $10^{\circ}$, for example if we rotate the 1st, or rotate it $\alpha^{\circ}$, then again we have a root system, because all of the four conditions stated there remain invariant.

But why we just consider only $45^{\circ}$ rotation?

I think there is something very simple, which I can not see, but I can not find it.

edit: In comments, it is mentioned that one is related to $A_1\times A_1$, and the other one is related to $D_2$. My question is why we can not consider $10^{\circ}$ rotation? Why it is not working?

edit: I think if we rotate the root system $A_1\times A_1$ exactly $45^{\circ}$ then we have the root system of $D_2$. Am I mistaken? What do the diagrams that differ by rotation (one is obtained by the rotation of the other one) have to do with each other? What is the difference between the root system of $A_1\times A_1$ and its $10^{\circ}$-rotation? What is the difference between the root system of $A_1\times A_1$ and its $45^{\circ}$-rotation (which I think is $D_2$)?

edit: Why do we just consider the system root of $A_1\times A_1$, and its $45^{\circ}$-rotation? Why we do not consider the system root of $A_1\times A_1$, and its $10^{\circ}$-rotation, and its $20^{\circ}$-rotation? What is special about $0^{\circ}$-rotation of the system root of $A_1\times A_1$, and $45^{\circ}$-rotation of the system root of $A_1\times A_1$?

Tireless and hardworking
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  • Please always include the link you are talking about. The other one is $D_2$, which is indeed isomorphic to $A_1\times A_1$. So no problem. – Dietrich Burde Dec 19 '21 at 20:03
  • @DietrichBurde I add the link to my question. My question is why we can not consider $10^{\circ}$ rotation? Why it is not working? Your comment is not an answer to my question. I was asking why other rotations do not work? Your comment is about a special rotation ($45^{\circ}$) that works. – Tireless and hardworking Dec 19 '21 at 20:12
  • You can rotate this picture as much as you want. Or do I misunderstand something? If the root system has rank $2$ (so in the plane) and the Dynkin diagram is not connected, we always have the root system $A_1\times A_1$, which we can rotate as much as we like. – Dietrich Burde Dec 19 '21 at 20:23
  • @DietrichBurde I think if we rotate the root system $A_1\times A_1$ exactly $45^{\circ}$ then we have the root system of $D_2$. Am I mistaken? What do the diagrams that differ by rotation (one is obtained by the rotation of the other one) have to do with each other? What is the difference between the root system of $A_1\times A_1$ and its $10^{\circ}$-rotation? What is the difference between the root system of $A_1\times A_1$ and its $45^{\circ}$-rotation (which I think is $D_2$)? – Tireless and hardworking Dec 19 '21 at 20:34
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    The simple roots are vectors in $\Bbb R^l$, where $l$ is the rank. By choosing a different basis of $\Bbb R^2$, the pictures can be rotated. – Dietrich Burde Dec 19 '21 at 20:37
  • @DietrichBurde Why do we just consider the system root of $A_1\times A_1$, and its $45^{\circ}$-rotation? Why we do not consider the system root of $A_1\times A_1$, and its $10^{\circ}$-rotation, and its $20^{\circ}$-rotation? What is special about $0^{\circ}$-rotation of the system root of $A_1\times A_1$, and $45^{\circ}$-rotation of the system root of $A_1\times A_1$? – Tireless and hardworking Dec 19 '21 at 20:43
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    I think everyone is trying to tell you that your skepticism is right, there is nothing special about this 45° rotation. It's a bit funny of wikipedia to handle these two separately. They are really the same root system, and so would be all infinitely many other rotations of it. – Torsten Schoeneberg Dec 19 '21 at 21:50
  • @TorstenSchoeneberg +1 Thank you. Your brief comment is exactly what I wanted to hear.

    I think I explained my questions very well in "edit"s, and I think your comment is a perfect answer to it.

     – Tireless and hardworking Dec 20 '21 at 14:10

1 Answers1

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The two important features of a root system are the angles between, and the relative lengths. Rotating the whole thing does not change either.

From my viewpoint, the $D_2$ root system is literally the same as (meaning isomorphic to) the $A_1\times A_1$. Indeed, the corresponding Lie algebras are isomorphic.

It's true that usually the simple roots are aligned in traditional ways, but that is mathematically inessential.

paul garrett
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