Math 54 Transition graphs and subshifts November 6, 04. Transition graphs Let I be a closed interval in the real line. Suppose F : I I is function. Let I 0, I,..., I N be N closed subintervals in I with pairwise disjoint interiors. Assume that F is continuous on each I j. 4 0 4 Figure.. An example of a function F. Let I j = [j, j + ], for j = 0,,,. The transition graph of F is a graph G formed by () one vertex for each subinterval I j, labelled by the index j, and () an arrow i j if F (I i ) I j. Figure.. The transition graph for the example in Figure.. Theorem.. Suppose F is as above. Let p = v 0 v v n be an infinitely repeating path of vertices in the graph G. Suppose that
() n is the smallest period of p (so p is not of the form v 0 v v m for some m < n), and () the endpoints of the interval I v0 do not have itinerary p. Then there exists a point x I v0 of exact period n for F with the itinerary p. Before giving the proof of Theorem., let s see how it applies for the example in Figure.. Consider first the repeating sequence p = =, with period n =. Look at the itineraries of the endpoints of I = [, ]. Note that x = lies in a cycle of period. The itinerary of x = is not well defined, but it can be any sequence of the following form: x 0 x x x = (0 or )( or )(0 or )( or ) That is, every even entry must be a 0 or, and every odd entry must be a or a. In particular, our p = cannot be the itinerary of x =. The itinerary of x = is more straightforward: it is x 0 x x x = ( or )0 so it cannot coincide with p. Therefore, this function F has a fixed point inside I. Of course, we can see immediately from the graph of F that it has a fixed point in I. For a less straightforward example, consider p = 0 of period 4. Again we begin by computing the itineraries of the endpoints of I 0. The itinerary of x = 0 begins with 0, so it cannot coincide with p. We have already computed the itinerary of x =. Interestingly, our given p is one of the possible itineraries for the point x =. So unfortunately, we cannot deduce from Theorem. whether or not there exists a point of period 4 in I 0. On the other hand, consider the sequence p = 0. The Theorem does allow us to conclude that there is a point of period exactly 4 with this itinerary in I 0. The graphs of the nd and 4th iterates of F in Figure.. Note that the 4th iterate has exactly one fixed point in I 0 in addition to its endpoint (which has exact period ). The itinerary of that point must be 0. We see that there is no point of period exactly 4 with itinerary 0. 4 4 0 4 0 4 Figure.. The nd and 4th iterates of the function in Figure.. The points x = and x = are in a cycle of period. There is a point in I 0 that has period exactly 4 and itinerary 0. *** INSERT PROOF OF THEOREM. HERE ***
. Transition matrices and the subshift The transition matrix is an N-by-N matrix T that contains the same information as the transition graph, only presented in a different format. If we index the rows and columns by 0,,..., N, then ij-th entry in the matrix is a 0 or, where T ij = there is an arrow from i to j in G For the example in Figures. and., we have the matrix 0 0 0 (.) T = 0 0 0 0 0 0 0 Recall that the shift space on N symbols is Σ N = {0,,..., N } N, the set of all infinite sequences s = s 0 s s such that s i {0,,..., N } for all i. Recall that we defined a distance on Σ N by { 0 if s = t d(s, t) = if n = min{i : s n i t i } The subshift space associated to a transition matrix T is the set Or, in terms of the graph G, Σ T = {s 0 s s Σ N : T si s i+ = for all i}. Σ T = {sequences that are infinite paths in the graph}. Sometimes I will write Σ G for Σ T to emphasize the dependence on a graph. Figure. shows the simple transition graph I presented in class. For this example, the space Σ T is fairly easy to describe. It is: Σ T = {0, 0,, 0, 0, 00, 00, 000, 000,...} Figure.. Another transition graph. The subshift defined by a transition graph G or its transition matrix T is simply the shift map restricted to Σ T. That is, σ : Σ T Σ T defined by σ(s 0 s s ) = s s s. For the example of Figure., the sequence is fixed by σ, and the points 0 and 0 form a cycle of period. All other elements of Σ T are prefixed; after some number of iterates, they land on.
4 There is a definite advantage to encoding the data of G in matrix form, as the transition matrix T, as the following result shows. Theorem.. Let T be a transition matrix, and let σ : Σ T Σ T be the associated subshift. Then the number of fixed points of the n-th iterate σ n is equal to the trace of the matrix T n. Recall that the trace of a matrix is the sum of its diagonal entries. So for n = in the statement of the theorem, we have that tr T = #Fix(σ). Indeed, tr(t ) = T 00 + T + + T N,N, which must coincide with the number of fixed points of σ, by the definition of the transition matrix. For the example of Figure. with transition matrix (.), there is one sequence fixed by σ. The trace of T is also. Since the fixed point of σ is also fixed by σ, we see that there are no points of period exactly in Σ T. On the other hand, tr(t ) = 4, so there are four sequences in Σ T fixed by σ. These are {, 0, 0, 0}. Among these, one is the fixed point and the other three form a cycle of period. *** INSERT PROOF OF THEOREM. HERE ***. Chaotic subshifts Now we move to the question of when a subshift is chaotic. Recall that a dynamical system f : X X on a metric space X is chaotic if three conditions are satisfied: () periodic points are dense in X; () there exists a point x X with a dense orbit; and () f depends sensitively on initial conditions. Condition () means that there is a constant δ > 0 so that for every x X and every ε > 0, we can find a point y X and an iterate n N, so that d(x, y) < ε and d(f n x, f n y) δ. From the definitions, we may observe that the example in Figure. does not define a chaotic subshift. For example, this subshift fails to satisfy condition (). There are only three periodic points in Σ T, and none of these is close to the point s = 0. In fact, d(0, x) = / for each of the three periodic points x. So periodic points cannot be dense in this Σ T. By contrast, the next theorem will show that the example from Figure. is chaotic. Theorem.. Let G be a transition graph. Assume each vertex in G has at least one arrow pointing from it. Then the subshift σ : Σ G Σ G is chaotic if and only if G is irreducible and there exists a vertex with at least two arrows pointing from it. A transition graph is irreducible if for any pair of vertices i, j {0,,..., N }, there is a path from i to j. The examples from Figures. and. are both irreducible. In terms of the transition matrix T, the graph G is irreducible if and only if for each pair i, j, there is a power n so that (T n ) ij > 0. *** INSERT PROOF OF THEOREM. HERE ***
4. Transferring the information back to the original function It remains to relate these properties of the subshift back to properties of the original function F on an interval or union of intervals. Assume that X = I 0 I I N is a union of N closed intervals in R, with pairwise disjoint interiors. A differentiable function F is said to be expanding on X if there is a constant β > so that F (x) β for all x X. We also allow F to be non-differentiable at the endpoints of the intervals I j, as in Figure.. For that example in Figure., however, F is not expanding on all of X = I, since its slope is equal to on I 0 and I. But it is expanding on I I, where the slope is greater than in absolute value. Theorem 4.. Assume that F : X R is expanding. Assume also that F is one-to-one on each of the intervals I j. Let G be the transition graph for F with N vertices, and let σ : Σ G Σ G be the associated subshift. () For every sequence s Σ G, there is a unique point x X with itinerary s. From (), we see there is a well-defined map h : Σ G X sending s to this x. Let Λ G = h(σ G ) X. () The map h : Σ G Λ G induces a semiconjugacy from σ to F ; that is, F h = h σ. () If σ is chaotic on Σ G, then F is chaotic on Λ G. Though the function F from Figure. is not expanding, the third iterate of F is expanding. One can exploit this fact to deduce that Theorem 4. will apply also to this example. Therefore, this F is chaotic on the interval [0, ]. I leave it for you to think about why this might be true. There is only one key idea in the proof of Theorem 4. that you need to know. It is the construction of the point x with itinerary s. We have seen this before when studying the dynamics of Q c (x) = x + c with c <. For a given itinerary s = s 0 s s, look at the intersection I s0 F (I s ) F n (I sn ). Any point x in this intersection will satisfy F j (x) I sj for all j n. Since F is one-toone and continuous on each interval, this intersection will be a closed interval. Since F is expanding with expansion constant β >, this interval has length at most /β n times the length of the largest of the subintervals I j. In particular, as n, the infinite intersection will consist of a single point. That is our desired x. By construction, we have F n (x) I sn for each n 0. 4.. Tent map. Let s illustrate Theorem 4. with an important example, the tent map on the unit interval [0, ]. Recall that the tent map of slope is defined by { x if 0 x T (x) = x if x 5
6.0 0.8 0.6 0.4 0. 0.0 0. 0.4 0.6 0.8.0 Figure 4.. The tent map of slope. For the tent map, we define I 0 = [0, /] and I = [/, ]. Note that T is expanding, since T (x) = at all points where the derivative is defined. The subshift for T with this choice of intervals I 0, I is actually the full shift on symbols, σ : Σ Σ, where every transition 00, 0, 0, is allowable. We have seen in class that the full shift is chaotic. Theorem 4. then implies that T is chaotic on the interval [0, ]. Moreover, T is semiconjugate to σ. Note that this semi-conjugacy cannot be a conjugacy! There is no homeomorphism between Σ and the interval [0, ]. After all, Σ is a Cantor set (and so totally disconnected), while the interval is connected. But the itinerary map h : Σ [0, ] is a continuous projection onto the interval. Its failure to be one-to-one can be seen quite easily. Note that the critical point x c = / does not have a well-defined address. This point x c lies in both I 0 and I. So it is described by two different itineraries, 0000 and 000 Part of the content of Theorem 4. is that x c is the only point in [0, ] with either of these itineraries. Similarly, look at the preimages of x c, namely x 0 = /4 and x = /4. Each of these points also has two itineraries; for x 0, they are 00000 and 0000, while for x they are 0000 and 000. The critical point and its preimages under iteration of T are the only points with multiple itineraries. Excluding these points and itineraries, h becomes a conjugacy. 4.. Quadratic examples. As another example, recall that the quadratic function Q (x) = x on [, ] is conjugate to the tent map T. We gave the explicit conjugating function, expressed in terms of the sine function, in class. It follows that Q is chaotic on [, ] (and semiconjugate to the full shift on symbols), as we have mentioned many times throughout the quarter. But Theorem 4. gives us a proof of this fact. An even more interesting example is the quadratic polynomial Q c (x) = x + c, where c = c is chosen so that the critical point x = 0 has period exactly. Thus c.755 lies in
the period- window of the bifurcation diagram. Since the derivative of Q c will be 0 at 0, this cycle of period is attracting. For this c, note that Q c (0) = c and Q c(0) = c +c = c (since Q c(0) = 0). The graph of Q c on the interval I = [c, c] is shown in Figure 4.. 7.0 0.5 -.5 -.0-0.5 0.5.0-0.5 -.0 -.5 Figure 4.. The graph of Q c with c.755, on the interval I = [c, c] Set I 0 = [c, 0] and I = [0, c]. The transition matrix for Q c : I I is ( ) T = 0 You can check that the subshift σ : Σ T Σ T is chaotic. But note that Q c is not chaotic on I. After all, Q c has an attracting cycle of period. So there is a small interval around 0 that follows the orbit of 0. More precisely, the iterates (Q c) n (x) 0 as n for all x in a small neighborhood of 0. In particular, there cannot be a dense orbit in I and the periodic points cannot be dense. On the other hand, the subshift σ : Σ T Σ T is still useful for understanding the dynamics of Q c : I I. It is still the case that every allowable sequence s Σ T must be the itinerary of some point x I. This follows from the same ideas that go into proving Theorem.. What fails is that the point is not uniquely determined by its itinerary because of the presence of the attracting cycle. For example, there is a small number δ > 0 so that all x in the interval ( δ, 0) have the same itinerary, namely, 00, and all x in the interval (0, δ) have the itinerary 000. The critical point x = 0 has an ambiguous address and therefore the ambiguous itinerary of (0 or )0(0 or )0(0 or )0. But only two possibilities actually lie in Σ T, since is not an allowable transition. Therefore, the only allowable itineraries for the critical point are (0 or )000.