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Hello all,

I want to prove that any flow on the following tree must have an infinite energy.

The structure of the graph is (taken from R.Lyons and Y.Peres book)

"We’ll construct a tree $T$ embedded in the upper half plane with o at the origin. We’ll have $|T_n| = 2^n$, but we’ll connect them in a funny way. List $T_n$ in clockwise order as $(x^n_1 , . . . , x^n_{2^n})$. Let $x^n_k$ have $1$ child if $k \leq 2^{n−1}$ and $3$ children otherwise."

Any idea?

Thanks in advance!

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Well, this tree is recurrent, so of course any flow must have infinite energy... –  Ori Gurel-Gurevich Jul 19 '12 at 18:58
    
Ori, i want to prove that this tree is recurrent using the fact that any flow must have... Lee, you right. First, it should be $|T_n| = 2^n$. It means that the cardinality of the $n$th level of the tree is equal to $2^n$ ($2^n$ vertices). Flow is an anti-symmetric function for which the divergence is not zero at the root and zero elsewhere. The energy is, of course, defined by the inner product on $L^2(E)$ (edges). –  StarDust Jul 19 '12 at 19:24
    
Have a look at mathoverflow.net/questions/102283/… –  Anthony Quas Jul 19 '12 at 20:26
    
Thank you! The truth is that I already saw this thread, but again, I wish to prove the recurrences property directly by showing that any flow must have infinite energy. –  StarDust Jul 19 '12 at 20:56
    
StarDust: I understood that, should've added a smiley ;-) To the point, the 3-1 tree "ends" in countably many infinite paths. If the flow assigns a positive quantity to any of these paths it yields infinite energy. –  Ori Gurel-Gurevich Jul 19 '12 at 21:44

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