What does $\mathbb{R}∗\mathbb{R}$ mean? I'm sure this has been asked before, but I do not know how to search for notations in past questions.
Question about Notation
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1Thinking back, I think Norbert is right. I'm pretty sure I've seen this used for either free product or abelian free product. – 2011-12-28
3 Answers
In group theory, $G ∗ H$ is defined as the free product of $G$ and $H$. It is an operation that constructs a new group which contains both $G$ and $H$ as subgroups, since it is generated by the elements of these groups. For definition, maybe the following link can be of help: http://mathworld.wolfram.com/FreeProduct.html.
Hatcher uses $X*Y$ to denote the join of two spaces, that is, the quotient of $X \times Y \times I$ by the identifications $(x, y_1, 0)$ with $ (x, y_2, 0)$ and $(x_1, y, 1)$ with $ (x_2, y, 1)$. This is the space of all line segments joining points of $X$ with points of $Y$. See page 9 of Hatcher's book on algebraic topology for more information.
If $X$ and $Y$ are closed intervals, the cube $X \times Y \times I$ gets collapsed to a tetrahedron. In the case of $X=Y=\mathbb{R}$, I guess we get an "infinite tetrahedron".
This is not an answer to the question. Since it is not easy to search for mathematical notation, perhaps this can help to confirm where the OP saw the notation. I am making this CW - feel free other occurrences from math.SE which may be relevant.
Searching for "\mathbb Z \ast \mathbb Z" site:math.stackexchange.com lead me to Georges Elencwajg's comment
Dear Akhil, I don't know about the existence of a purely algebraic proof. What I know is that until a few years ago there was no purely algebraic computation of the algebraic fundamental group of the projective complex line minus three points (namely $\pi_1^{alg}(\mathbb P^1_{\mathbb C} \setminus \{0,1, \infty\})=\widehat{\mathbb Z\ast \mathbb Z}$).
Searching for "\mathbb Z * \mathbb Z \to \mathbb Z \times \mathbb Z" site:math.stackexchange.com gives this question:
The first is, Is $i_*: \pi_1(S^1 \vee S^1) \to \pi_1(S^1 \times S^1)$ injective? My intuition is that no, this is not injective because $\pi_1(S^1 \vee S^1) = \mathbb{Z} * \mathbb{Z}$, the free group on two generators and $\pi_1 (S^1 \times S^1) = \mathbb{Z}\times \mathbb{Z}$. However, I am not sure if this is in fact true and I am trying to figure out the best way to go about showing it.
And from an answer to the same question:
Consider a homomorphism $f: \mathbb{Z} \ast \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$. If $a$ and $b$ are the generators of $\mathbb{Z} \ast \mathbb{Z}$ then consider what $ab$ and $ba$ map to:
$f(ab) = f(a)f(b) = f(b) f(a) = f(ba)$
In both cases it seems to denote free product.