Invariant element integrand

[Tio Barnabe]

Let $k$ be a Lorentz four-vector. The integrand $d^4k$ is the same as $d^3 k \ dk_0$. Why is this true?
$k_0$ is the first component of $k$. So how are we allowed for equating the two integrands?

Just to add context, this situation happens during the construction of the integral for the Klein-Gordon field solution "$\varphi(x)$".

 

[PeterDonis]

What does $d^4 k$ mean if $k$ is a four-vector?

(You should be able to answer this easily if you know what $d^3 k$ means if $k$ is a Euclidean 3-vector.)

 

[Tio Barnabe]

What does $d^4 k$ mean if $k$ is a four-vector?

The infinitesimal volume in the vector space spanned by $k$?

Would it be useful to consider wedge product here? Or is it unnecessary?

 

[Tio Barnabe]

Uhuuuul, I think I realized. $d^4 k = dk_0 \wedge dk_1 \wedge dk_2 \wedge dk_3$. Is it?

If so, then calling the remaining volume element of $d^3 k \ dk_0$ is a mistake of my book, because we should discriminate between $k = (k_0, k_1, k_2, k_3)$ and, say, $\bf{k}$ $= (k_1, k_2, k_3)$.

 

[Nugatory]

If so, then calling the remaining volume element of $d^3 k \ dk_0$ is a mistake of my book, because we should discriminate between $k = (k_0, k_1, k_2, k_3)$ and, say, $\bf{k}$ $= (k_1, k_2, k_3)$.

It's not a mistake, it's just that you're being expected to interpret the notation properly based on the context.