# A symmetric problem has an asymmetric solution - why?

• B
• Omega0
Omega0
TL;DR Summary
A set of two symmetric non-linear algebraic equations has an asymmetric solution. What is the deeper reason behind this:
Hello,

please let me split me split my question into 3 blocks. The first: The problem and the solution. The second: The question. The third: Maybe weird thoughts about about a similar problem.

The problem and the solution

\begin{align*} 6^x+6^y &= 42 \\ x+y &= 3 \end{align*}

This is solved by ##x=1, y=2## or ##y=2, x=1##.

The questions I have

First, it is 100% clear that if a solution for x and y is found, they need to be exchangeable. The problem is absolutely symmetric.
And this is what puzzles me. What is the intrinsic reason that the solution isn't symmetric, meaning ##x = y##? What mathematics is behind this?

Please note that the first equation is nonlinear. If the set of equations would be symmetric and linear, there wouldn't be any solution or an infinite amount of solutions.
Let me tell you about my thoughts: It remembers me around the Karman Vortex Street. With the difference that we have here only 2 points in the solution space. The problem to get the Karman Vortex Street is nevertheless perfectly symmetric,, mathematically. The equations are non-linear. If you simulate them you'll find out that it takes some time to get the oscillation - but it will come!
In nature it is totally clear: We have chaotic movement etc., no problem - but for mathematics, I thought we wouldn't have a broken symmetry for such a simple problem. The problem above appears to me like a broken symmetry.

Jens

Omega0 said:
What is the intrinsic reason that the solution isn't symmetric, meaning ##x = y##?
That is not what is meant as "symmetric" in the XY plane. A set does not have to be a subset of the line x=y to be symmetric. The definition is: The set ##S## is symmetric if ##(a,b) \in S## implies ##(b,a)\in S##.

Lnewqban
FactChecker said:
That is not what is meant as "symmetric" in the XY plane. A set does not have to be a subset of the line x=y to be symmetric. The definition is: The set ##S## is symmetric if ##(a,b) \in S## implies ##(b,a)\in S##.
Okay, I should have said invariant under an exchange of the variables? This problem is not about lines or so, variable names are just dust in the wind.

Omega0 said:
Okay, I should have said invariant under an exchange of the variables? This problem is not about lines or so, variable names are just dust in the wind.
Your problem is invariant under the exchange of variables, and so are the solutions.

fresh_42 said:
Your problem is invariant under the exchange of variables, and so are the solutions.
Yes, I know, but what is behind this? That is my question. Again: A linear set of equations which is invariant under exchange of the variables has an infinite number of solutions.
As soon as we have a non-linearity, the solution space seems to collapse. I would like to know why and how.
Thanks.

The linear equations are not independent of each other.

Assume we have a system of equations ##f_1(x_1,\ldots,x_n)=0,\ldots, f_m(x_1,\ldots,x_n)=0## and a permutation ##\pi \in \operatorname{Sym}(n)## such that ##f_1(\pi(x_1,\ldots,x_n))=0,\ldots, f_m(\pi(x_1,\ldots,x_n))=0.## This means that for any solution ##x_1=a_1,\ldots,x_n=a_n,## i.e. ##f_1(a_1,\ldots,a_n)=0,\ldots, f_m(a_1,\ldots,a_n)=0## we also have ##f_1(\pi(a_1,\ldots,a_n))=0,\ldots, f_m(\pi(a_1,\ldots,a_n))=0.##

In words: If we can distinguish between variables in a solution, they cannot be indistinguishable in the problem statement, because otherwise solving the system would be a method to distinguish them.

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Your question is good. To simplify it, you only need one equation. Suppose x+y=3. That is symmetric in the variables and the solution set is a line, y=3-x, which is not the same as y=x. The solution set is still symmetric with respect to the line y=x. Here is the plot. You can see that for any point (x=a,y=b) on the solution line the point (x=b, y=a) is also on the solution line. For instance, (x=0, y=3) and (x=3, y=0) are both on the solution line.

Lnewqban
An even simpler case would be this: if x=3 then y=0 and if x=0 then y=3. This statement is symmetric in the variables x and y, but in no case is y=x. The solution set is {(x=0, y=3), (x=3, y=0)}. Symmetry does not mean equality.

Lnewqban
FactChecker said:
Your question is good. To simplify it, you only need one equation. Suppose x+y=3. That is symmetric in the variables and the solution set is a line, y=3-x, which is not the same as y=x. The solution set is still symmetric with respect to the line y=x. Here is the plot. You can see that for any point (x=a,y=b) on the solution line the point (x=b, y=a) is also on the solution line. For instance, (x=0, y=3) and (x=3, y=0) are both on the solution line.
Thanks. Unfortunately, this example doesn't fit because in the case that you have only ##x+y=3## you have a function which is invariant under variable exchange but it is exactly the case I mentioned in my post: Now you have an unlimited number of solutions. This doesn't count. I am talking about a set of equations.
Your second equation now would fulfill indeed the need to be invariant again but the only function working in the linear case is indeed ##y=x##. Unfortunately, this means that the solution is the same for x and y, exactly what I mentioned, or say expected. You can't find a linear set of equations which is invariant under variable change and which also has a finite number of solutions which is not the same.
FactChecker said:
An even simpler case would be this: if x=3 then y=0 and if x=0 then y=3. This statement is symmetric in the variables x and y, but in no case is y=x. The solution set is {(x=0, y=3), (x=3, y=0)}. Symmetry does not mean equality.
This is not correct because your set of equations is not invariant under variable exchange. You can see this writing:
\begin{align*} x+0y &= 3\\ 0x+y &= 0 \end{align*}

Or should I read it as a case distinction? This isn't a set of linear functions being invariant under variable exchange.

fresh_42 said:
Your problem is invariant under the exchange of variables, and so are the solutions.
Yes, this is clear. I am quite sure that for a set of equations which is invariant under the exchange of all variables there is something special happening. Like we know from tensor symmetries etc. I would like to understand, what - and naively expected an equality of the variables themselves. Perhaps I even learned that and just forgot it

Omega0 said:
Yes, this is clear. I am quite sure that for a set of equations which is invariant under the exchange of all variables there is something special happening. Like we know from tensor symmetries etc. I would like to understand, what - and naively expected an equality of the variables themselves. Perhaps I even learned that and just forgot it
I'm not sure what you are aiming at. Maybe this summary helps:
https://www.physicsforums.com/insights/when-lie-groups-became-physics/
or the hard-core version (p.235-257):

It is the beginning of Lie theory in physics and the rigorous way to consider symmetries. Picking arbitrary symmetric equations and talking about them won't get you very far.

Omega0 said:
Thanks. Unfortunately, this example doesn't fit because in the case that you have only ##x+y=3## you have a function which is invariant under variable exchange but it is exactly the case I mentioned in my post: Now you have an unlimited number of solutions. This doesn't count. I am talking about a set of equations.
You can add equations if you want to unnecessarily complicate things. Your confusion is that you think symmetry is the same as equality. That can be cleared up by looking at a simple, single function.
Omega0 said:
This is not correct because your set of equations is not invariant under variable exchange. You can see this writing:
\begin{align*} x+0y &= 3\\ 0x+y &= 0 \end{align*}

Or should I read it as a case distinction? This isn't a set of linear functions being invariant under variable exchange.
I do not mean for any linear functions to be involved at all. The only values defined are for the two values of x, x=3 and x=0. The definition is symmetric in the variables x and y, yet x and y are never equal for any of the two defined points, (3,0) and (0,3).

FactChecker said:
You can add equations if you want to unnecessarily complicate things.
This is not "to unnecessarily complicate things" but what my original post was about: A set of equations...
FactChecker said:
Your confusion is that you think symmetry is the same as equality. That can be cleared up by looking at a simple, single function.
Well, not sure how confused it is... you posted an example where the solution is neccessarily ##x=y## which is perfectly symmetric, right? See perhaps the definition, if it helps.
FactChecker said:
I do not mean for any linear functions to be involved at all. The only values defined are for the two values of x, x=3 and x=0. The definition is symmetric in the variables x and y, yet x and y are never equal for any of the two defined points, (3,0) and (0,3).
Okay, but how would you call a function which is not linear? See my post. It was about the idea that it has to do with functions which are neccesarily not linear. Also I would like to mention that the set of "my functions" is defined on ##\mathbb{C}##, where you are speaking about the set ##\{3,0\}##.

fresh_42 said:
I'm not sure what you are aiming at.
About solving non-linear equation sets, I would say - and the solution space.
fresh_42 said:
It is the beginning of Lie theory in physics and the rigorous way to consider symmetries. Picking arbitrary symmetric equations and talking about them won't get you very far.
Okay, thank you but I am not convinced that Lie groups are the right choice for non-linear problems?

Omega0 said:
Okay, thank you but I am not convinced that Lie groups are the right choice for non-linear problems?
Then it's good that Noether hasn't had these prejudices.

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Omega0 said:
This is not "to unnecessarily complicate things" but what my original post was about: A set of equations...
Having a set of equations is irrelevant to your basic misconception.
Omega0 said:
Well, not sure how confused it is... you posted an example where the solution is neccessarily ##x=y## which is perfectly symmetric, right?
No. None of my examples have a solution that forces ##y=x##.
Omega0 said:
See perhaps the definition, if it helps.
Ok. Your question was about symmetric solution sets. In particular, you have a solution set that is symmetric with respect to the line ##y=x##.
Omega0 said:
Okay, but how would you call a function which is not linear? See my post. It was about the idea that it has to do with functions which are neccesarily not linear.
I don't think that linearity necessarily has anything to do with this. The circle ##x^2+y^2=1## has a solution set that is symmetric with respect to the line ##y=x##, yet it is never true that ##y=x## for any of the solution points.
Omega0 said:
Also I would like to mention that the set of "my functions" is defined on ##\mathbb{C}##, where you are speaking about the set ##\{3,0\}##.
I don't see the relevance of that.

Omega0 said:
About solving non-linear equation sets, I would say - and the solution space.
Let me translate this. What is non-linearity? We can only handle equations of such a generality if we have a positive description. The most convenient set of functions that are not linear are functions that can be expressed by power series. Having complex variables, means we speak about analytical complex functions. Every set of solutions ##S=\{\vec{z}\,|\,\vec{F}(\vec{z})=0\, , \,F_j \text{ analytic }\}## can be interpreted as the solution of an IVP. Asking for symmetries in ##S## is asking for symmetries on the IVP, and this is Lie theory.

I don't think you will find a better answer than the calculus of variations.

Omega0 said:
This is solved by ##x=1, y=2## or ##y=2, x=1##.

And this is what puzzles me. What is the intrinsic reason that the solution isn't symmetric,
But the solution set is symmetric wrt the line y=x.
Omega0 said:
meaning ##x = y##?
That is not what a symmetric solution set means.

FactChecker said:
Having a set of equations is irrelevant to your basic misconception.
No, again, it is no misconception if you understand my setup. Nevertheless...
FactChecker said:
No. None of my examples have a solution that forces ##y=x##.
I believe I got something wrong, indeed. You said, that in 2D a symmetric function is symmetric to the line ##x=y## which I didn't valued. Sorry, but that's the point, visually the contour lines of a non-linear function being symmetric are symmetric to ##x=y## in 2D. The solution in my example (or other ones, like a circle), will be sort of ##(a,b)=(b,a)## - but ##a=b## is possible, which isn't interesting for me here.
FactChecker said:
Ok. Your question was about symmetric solution sets. In particular, you have a solution set that is symmetric with respect to the line ##y=x##.
Yes, true, as said above, I didn't appreciate this, my fault.
FactChecker said:
I don't think that linearity necessarily has anything to do with this.
Did I say that? I mean, from the very beginning, that we need at least one non-linear equation in our set of equations, where the solution set is finite, to have not an equality of the solutions.

The simplest example is of the following type:
\begin{align*} x\cdot y &= \frac{9}{100} \\ x+y &=1 \end{align*}
FactChecker said:
The circle ##x^2+y^2=1## has a solution set that is symmetric with respect to the line ##y=x##, yet it is never true that ##y=x## for any of the solution points.
Nope, it is true for the both solution points... the solutions are ##(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})##
and ##(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})##

As soon as you take a symmetric function like ##x+y=1## you'll have instead something like the more interesting solution ##(1,0)## and ##(0,1)## (which obviously remembers about group theory).
FactChecker said:
I don't see the relevance of that.
Well, I think it is nice to think about symmetries and it's applications or results. It is nice to see in such examples how symmetries are projected to the results.
My basic thoughts about "symmetry breaking" may have sounded too dramatic, but I believe it is a good idea to think also about easy cases. There is a lot hidden information in apparently easy mathematics.

Probably, my basic thoughts and it's mathematical formulation haven't been too precise, sorry about that.

FactChecker
fresh_42 said:
Let me translate this. What is non-linearity? We can only handle equations of such a generality if we have a positive description. The most convenient set of functions that are not linear are functions that can be expressed by power series. Having complex variables, means we speak about analytical complex functions. Every set of solutions ##S=\{\vec{z}\,|\,\vec{F}(\vec{z})=0\, , \,F_j \text{ analytic }\}## can be interpreted as the solution of an IVP. Asking for symmetries in ##S## is asking for symmetries on the IVP, and this is Lie theory.

I don't think you will find a better answer than the calculus of variations.
Well, I have another feeling... what about the following mathematical problem:

Given the following equation:

\begin{align*} 6^x+6y &= 42 \\ x+y &= a \end{align*}

Find the solutions for ##a \in \mathbf{N}## so that ##x## and ##y## are natural numbers.

This is algebra but has nothing to do with variational calculus, I would say. The question is not to find any extreme values, for example. It is about finding a very specific solution - just in this case.
My basic set of equations in the post have been formulated so that the solution is clearly defined. The question above is way harder.

Not variational formulation, Lie algebra or so will help. Or does it? Please let me know.

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Omega0 said:
Nope, it is true for the both solution points... the solutions are ##(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})##
and ##(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})##
Oops! I stand corrected. Sorry about that.

Omega0

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