# Why write $\mathbb R\times \mathbb R^3\to\mathbb R^3$ instead of $\mathbb R^4\to\mathbb R^3$?

I often see this notation for a vector field $$\mathbf A:\mathbb R\times \mathbb R^3\to\mathbb R^3$$

But isn't $\mathbb R\times \mathbb R^3=\mathbb R^4$, right?

So, is there any advantages using $\mathbb R\times \mathbb R^3\rightarrow\mathbb R^3$ instead of $\mathbb{R}^4\rightarrow \mathbb{R}^3$?

• It's to emphasize one particular variable, which, when specified, yields a function from $\mathbb{R}^3\to\mathbb{R}^3$. – quasi Jul 7 '17 at 9:40
• In addition, technically $\mathbb{R} \times \mathbb{R}^3 \cong \mathbb{R}^4$. – Joe Johnson 126 Jul 7 '17 at 9:48

## 1 Answer

The two sets are not equal, no. The first consists of pairs of the form $$(t, (x, y, z))$$ while the second contains elements of the form $$(t, x, y, z).$$ With that notation, there's a pretty natural isomorphism between the sets, as you've guessed, and mathematicians generally get pretty glib about saying they're "the same".

Why write the first way? Frequently, a vector field (the wind, for instance) is time-varying, and at each time $t$ and location $(x,y, z)$ there's a wind velocity. You'll frequently see that velocity denoted $F(t, v)$, where $v$ denotes a point in 3-space. You might want to talk about the field at one particular point $v_0$ changes over time, and if you like Leibniz notation, this becomes $$\frac{\partial F}{\partial t}(t, v_0).$$ You might also want to talk about how the field changes with respect to position, at a fixed moment, and this would be written $$\frac{\partial F}{\partial v}(t_0, v).$$ It's true that you could write, in the $\Bbb R^4$ version, something like $$\frac{\partial F}{\partial xyz}(t_0, x,y,z),$$ but that's often cumbersome.

Anyhow, the reason it's written that way is often just to make the naming of variables and their groupings easier to understand. Imagine if you did it the other way, but wrote $F(x, y, z, w)$ to indicate the wind speed at time $x$ and position $(y, z, w)$ ... that'd seem really awkward, because $x$ is not a natural name for a time coordinate, etc.

• By some definitions (i.e., the inductive definition of $\mathbb R^n$ as $\mathbb R^{n+1} = \mathbb R\times \mathbb R^n$), the sets $\mathbb R^4$ and $\mathbb R\times\mathbb R^3$ are actually the same. – 5xum Jul 7 '17 at 9:54
• I have seen $n$-tuples defined recursively as follows: $(x,y)=\left\{x,\left\{x,y\right\}\right\}$, $(x,y,z)=\left\{x,\left\{x,y\right\},\left\{x,y,z\right\}\right\}$, etc. Under this definition, John is right in saying that $\mathbb{R}^{n+1} \neq \mathbb{R}\times \mathbb{R}^n$. – user1892304 Jul 7 '17 at 10:07
• I agree with both of you. :) I should have made clear my "definition" of $\Bbb R^k$. I actually like to think of $\Bbb R^k$ as "functions from $\{1, 2, 3, \ldots, k\}$ to $\Bbb R$", because this makes the generalization of the standard inner product to the inner product on $L^2$ functions seem perfectly natural. Finally, if you want to go with @5xum's view, then you don't get that $\Bbb R^3 \times \Bbb R = \Bbb R^4$, which makes my answer useful in the obvious dual case of a field $F(v, t)$. :) – John Hughes Jul 7 '17 at 14:43