Baxter & Rennie at pag. 162 state the following theorem. 


Let $W$ be an $n$-dimensional $\mathbb Q$-Brownian motion and let $M\_t=(M\_1(t),...,M\_n(t))$ be an $n$-dimensional $\mathbb Q$-martingale process, which has volatility matrix $(\sigma\_{ij}(t))$, in that $dM\_j(t) = \Sigma\_i \sigma\_{ij}(t) dW\_i(t)$ and the matrix is non singular (for all $t$ with probability 1). Let $N\_t$ be any one-dimensional $\mathbb Q$-martingale. 


Then there exists an $n$-dimensional $F$-previsible process $\phi\_t = (\phi\_1(t), ... \phi\_n(t))$ such that $\int^T\_0 (\Sigma\_j \sigma\_{ij}(t) d\phi\_j(t))^2 dt <\infty$, and the martingale $N$ can be written as $N\_t = N\_0 + \Sigma\_j \int\_0^T \phi\_j(s) dM\_j(s)$. Further $\phi$ is (essentially) unique.


My question is: because each of the $M\_j(t)$ is a $\mathbb Q$-martingale, why can't I chose the $\phi\_j$ to write $N\_t=N\_0 + \int\_0^T \phi\_j(s) dM\_j(s)$ $\forall j$? In other words, what is the idea behind uniqueness in the $n$-dimensional case ($n>1$)?



n-factor martingale representation theorem
The statement and associated definitions in the book seem to gloss over an important assumption: the process $N\_t$ needs to be a martingale with respect to the measure $\mathbb{Q}$ *and* the filtration $\{\mathcal{F}\_t^W\}$ generated by the $n$-dimensional Brownian motion $W(t)$, i.e. $\mathcal{F}\_t^W = \sigma(W(s) : s \le t)$. That is, for each $t$, $N\_t$ needs to be $\mathcal{F}\_t^W$-measurable (i.e. the process is $\mathcal{F}\_t^W$-adapted), and for each $s < t$ we have $E[N\_t \mid \mathcal{F}\_s^W] = N\_s$.


This explains why your objection doesn't apply. I think your thought was that if you have a representation of a process $N\_t$ with respect to the Brownian motion $W$, you could apply the theorem again using the first coordinate $W\_1$, which is indeed a one-dimensional Brownian motion, in place of $W$, and get a different representation. But an $\mathcal{F}\_t^W$-martingale will typically not be an $\mathcal{F}\_t^{W\_1}$-martingale.


As a simple example, consider $N(t) = W\_2(t)$. Then $W\_2(t)$ is not measurable with respect to the $\sigma$-field $\mathcal{F}^{W\_1}\_t$: indeed, it is independent of this $\sigma$-field. So although $W\_2(t)$ is a martingale with respect to $\mathcal{F}\_t^W$, it is not an $\mathcal{F}\_t^{W\_1}$-martingale.


