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Suppose I take an Euclidean circle and identify antipodal points, then the inner points of the circle and its border are a model of the projective plane.

What would a straight line in this model mean?

How would a straight projective line in this model look like?

Or are there many alternatives?

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    I don't understand how this gives you a model of the projective plane. If you have a line that does not pass through the center of the circle, then the two intersections with the border are not antipodal (and so represent two different points); then this line intersects the line at infinity (the border) in two points? – 2017-02-01
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    No. The line you draw with two different points at infinity is only visually a line in te 2-D model om paper. It is not conceptually a line of the proj. plane. Therefore I asked: how does a straight line look like in the paper 2-D model. – 2017-02-02

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"Straight" does not make sense since the construction you describe is a topological construction. If you draw a line on the disk, yes it will looks like a straight usual line of $\mathbb R^2$ intersected with $D^2$ (so the endpoint will be identified : topogically it will be a circle).

But if you want to "imagine" what the lines on the projective space looks like, they are embedding on the projective space, and these embeddings will not be send lines of $\mathbb RP^2$ to straight lines.

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    I know this, but I wonder whether there is a possible interpretation for a straight line, and how a straight line would look like when not straight? – 2017-01-22
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    This depends on the embedding of $\mathbb RP^2$ you will choose. For example, if you take a line in $\mathbb RP^2$, your line will be included in $\mathbb RP^2 \backslash D$ where $D$ is a disk, and since $\mathbb RP^2 \backslash D \cong M$ where $M$ is a Moebius band, this gives you straight line of $\mathbb RP^2$ as "soul" of a Moebius band, or as a boundary of a disk in $M$. Maybe you are interested in the curvature of the projective plane ? It is the same as the curvature of $S^2$. Geodesics of $\mathbb RP^2$ are projections of geodesics of $S^2$, so straight line in the disk model. – 2017-01-22
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Think of the projective plane as half a sphere, border included, with antipodal points in the border circle identified. A projective line is then half a great circle with the points in the border circle identified. If you want to think of the "plane" model, think you are in R^3, project, say, the unit sphere into the x,y-plane.Then you can think of the projective plane as being obtained, via orthogonal projection, from a unit disk with antipodal border points identified. Projective lines are then the projections of the half circles mentioned above with the border points identified.

Hope my english makes it clear.

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    I forgot to mention a vital point. Half circles result from intersecting the half sphere with affine planes going through the centre of the sphere. But so does the border circle. So this border circle, antipodal points identified, is also a projective line. The same is true for the boundary of the disk model.You can forget about the half sphere and think of the sphere with antipodal points identified. Then projective lines are just great circles with antipodal points identified. that is, circles with antipodal points identified. there are bijections between the 3 models. – 2017-02-01