In a standard undergraduate course in linear algebra, one
learns that the only subspaces of the plane are the zero subspace,
the entire plane itself, and the (straight) lines through the origin.
This is easily implied from the definition of a subspace which requires
it to be closed under linear combinations and scalar products. However
we could not have reached this conclusion by just using the fact that a
subspace is also a **subgroup** of the plane. Consider the following
example.

**Example 1.** Let
*L = \{(x,y): x \text{and} y \text{are integers} \}*. Then *L* is
clearly a subgroup of the plane that is not a line through the origin or
the entire plane or the origin. Of course *L* is not a subspace since,
for example, it is not closed under scalar multiplication.

Noticing that the lattice *L* described in Example 1 is not a
connected subset of the plane, suppose that we impose the extra
restriction of connectedness on our subgroup. Can we conclude that
the subgroup is one of the special types mentioned above?
Although the situation here is not as obvious, the non-trivial
Example 2 at the end of the paper will show that the answer is still
``No''. An additional hypothesis is needed. In this paper we will show
the following result.

A subgroup must be either a straight line through the origin or the entire plane if either

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]:

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:

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.

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.

**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.

**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.

**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.

**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.

Long Island University/C.W.Post Campus

Brookville NY 11548 USA

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