ADDITIVE SUBGROUPS OF THE PLANE

or,

Next, we ask, if in (1) above, we replace connected with the stronger condition of arcwise connected, can we eliminate the additional hypothesis of containing a linear d-arc and still draw the same conclusion? Here the answer is ``yes'' (now allowing the trivial subgroup, i.e. the origin, as a possibility) and this is a consequence of the much broader result [6]:

Yamabe's Theorem

The reader unfamiliar with the concept of a Lie group need not be concerned since the only Lie group of interest in this paper is the plane, denoted by R^2. Applied to this group, the theorem affirms one's intuition about (additive) subgroups of the plane, namely:

Yamabe's Theorem in the Plane

Note that if g: [a,b] \rightarrow R^2 then, by virtue of the Inverse Function Theorem, the set \{g(t) = (x(t),y(t)): a < t < b\} will be a d-arc as defined above provided that x(t) is a differentiable function with continuous derivative and x'(t_0) \neq 0 for some t_0 \in (a,b). This conclusion also holds if x(t) is replaced by y(t) in the statement.

All subgroups of the plane will be assumed to be additive subgroups.

We can now state a special case of Yamabe's theorem in the plane.

Corollary

We point out that even in the case of the plane, Yamabe's theorem is not true if `arcwise connected' is replaced by `connected': see the example at the end of the paper. However, the additional hypothesis that the subgroup contains a d-arc is sufficient. We continue with two easy observations.

Observation 1

Proof. Suppose that H contains the non-vertical line L = \{(x,mx+b): x \in R\}. Looking at Figure 1, we see that A-B+P = C which is an arbitray point on the line L'. Here A = (x,mx+b), B = (0,b) and C = (x+c, m(x+c) + (d-mc)). The proof is even simpler if L is a vertical line.

Observation 2

Proof. As in the proof of Observation 1, points on the given line segment S can be subtracted to obtain a line segment S' which is parallel to S through the origin. Then additions of points on S' include the entire line containing S' and hence, by Observation 1, the line containing S.

We now come to the main result.

Proposition

Outline of proof. To prove (1) we argue as follows. If the subgroup H contains a nonlinear d-arc, then we will use additivity of H to show that it must contain an interval of the Y-axis and so must contain the entire Y-axis. From this it is easy to shwo that H must also contain the entire X-axis and hence the whole plane. (Additivity again!)

Proof of Proposition. Remark: The proof we give below is for the case that H contains a d-arc of the form \{(x,f(x)): a < x < c \}. A similar argument handles the case when the arc is of the form \{(f(x),x): a < x < c\}.

Claim: If f is continuous then f is linear if and only if g(x,y) = f(x)+f(y)-2f((x+y)/2) = 0.

Proof of claim: If f is linear then clearly g(x,y) = 0. On the other hand, suppose that g(x,y) = 0 for all x and y. In this case, the line segment joining the points A = (x,f(x)) and B = ((x+y)/2, f((x+y)/2)) = ((x+y)/2, (f(x)+f(y))/2) has the same slope as the line segment joining B and C = (y,f(y)). Therefore, for any values of x and y the three points A,B,C are collinear. Thus, by continuity of f, f is linear and the claim is shown.

Note. If the function g above is a constant function, then substituting x=y into the equation for g shows that g must actually be equal to the zero function: that is g is constant if and only if g=0.

We now return to the proof of the proposition.

(1) Choose real numbers x and y such that a < x,y < c. Then a < (x+y)/2 < c. Therefore the following three points lie in H (see Figure 2):
A = (x,f(x)), B = (y,f(y)), C = ((x+y)/2, f((x+y)/2)).
Since H is an additive group, the point D = A+B-2C = (0,g(x,y)) is also in H. Since f is non-linear, by the note above, g(x,y) = f(x)+f(y)-2f((x+y)/2) is not constant. Then, being a continuous function from R^2 to R that is not constant, the image of g must contain an interval (t_1,t_2), say, in R. This simply says that H contains the interval \{(0,r): t_1 < r < t_2\} of the Y-axis. It then follows from Observation 2 that H contains the entire Y-axis. Now for any x \in (a,c) we have (x,f(x)) \in H, and since H contains the entire Y-axis, H contains the point (x,f(x)) - (0,f(x)) = (x,0) for any x \in (a,c). Again, by Observation 2, H must now contain the X-axis. Since every point (s,t) in the plane is the sum of a point on the X-axis and a point on the Y-axis (because (s,t)=(s,0)+(0,t)), we have shown that H = R^2 if f is nonlinear.

(2) Now suppose that H contains a linear d-arc and a point P which is not on the line L containing this arc. Let L' be the line through P and parallel to L. By Observations 1,2 respectively, H contains L' and L. Consider the subgroup S = H \cap Y-axis. It is a connected subgroup of the y-axis and so must either be zero (the origin) or the entire Y-axis. This follows from Observation 2 because a connected subset of the real line is either a point, an interval, or the whole line itself. Now L and L' have different y-intercepts and so S must be the entire Y-axis. This implies that H contains all lines parallel to L (again by Observation 1) and this is the entire plane.

The following result is a special case of Yamabe's theorem in the plane.

Corollary

Proof. If the d-arc is non-linear then by Proposition (1) the subgroup must be the entire plane. If the d-arc is linear and the subgroup contains a point not on the line containing this d-arc, then, by Proposition (2), the subgroup must be the entire plane. If the subgroup contains no points not on the line, then the subgroup is just a straight line through the origin.

The following example (based on nontrivial results of F.B. Jones [3],[4]) shows that the hypothesis of `arcwise connected' in Yamabe's Theorem cannot be replaced by `connected'.

Example 2. Although F.B. Jones does not specifically address Yamabe's Theorem, it is easy to make the connection from his work. Recall that a Hamel basis of R is a set of elements of R that is linearly independent over the rational numbers and is maximal with respect to this property. Jones uses a Hamel basis of R to define a function f:R \rightarrow R with the following properties.

Here Gph(f) = \{(x,f(x)): x \in R\} denotes the graph of f.

The set Gph(f) defined above is clearly a nontrivial, connected, additive subgroup of the plane. By (c) it is neither a straight line through the origin nor the entire plane. We refer the reader to [4] for the definition of f and the proof that it actually satisfies (a)-(c) above.

References

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