Mrthdrb

I was reading a research article on quantum computing and didn't understand the tensor notations for the unitary operations. The article defined two controlled gates.

Let $U_2^m$ be a $2^m times 2^m$ unitary matrix, $I_2^m$ be a $2^m times 2^m$ identity matrix. Then, controlled gates $C_n^j(U_2^m)$ and $V_n^j(U_2^m)$ with $n$ control qubits and $m$ target qubits are defined by $$ C_n^j(U_2^m)=(|jrangle langle j|) otimes U_2^m+ sum_i=0,i neq j^2^n-1((|irangle langle i| otimes I_2^m$$

$$ V_n^j(U_2^m) = U_2^m otimes (|jrangle langle j|) + sum_i=0,i neq j^2^n-1( I_2^m otimes (|irangle langle i| ))$$
Then they say that $C_2^j(X)$ and $V_2^j(X) $are toffoli gates.
Can someone explain the equations that are given
and how does this special case be a Toffoli?

Here $i$ and $j$ are bit strings of size $n$. Correspondingly, $|irangle$, $|jrangle$ are some basis vectors in $2^n$-dimensional space, that corresponds to $n$-qubit register.

Those controlled operations $C$ and $V$ act on $(n+m)$-qubit space. You can consider first $n$ qubits as control register and last $m$ qubits as target register. Now, $C_n^j(U_2^m)$ applies unitary operation $U_2^m$ on the target register if control register is in the state $|jrangle$ and applies $I_2^m$ (i.e. do nothing) otherwise. You can see this by applying $C_n^j(U_2^m)$ on some vector $|xrangle|yrangle$ from the $(n+m)$-qubit space, where $x$ is some $n$-bit string:

$$
C_n^j(U_2^m) |xrangle|yrangle = (|jrangle langle j|xrangle) otimes U_2^m |yrangle+ sum_i=0,i neq j^2^n-1((|irangle langle i| x rangle) otimes |yrangle)
$$

Here $|irangle langle i|xrangle = 0$ if $xneq i$ and it equals $|irangle$ if $x=i$.
Hence
$$C_n^j(U_2^m) |xrangle|yrangle = |jrangle otimes U_2^m |yrangle + 0 = |xrangle otimes U_2^m |yrangle ~~textif~~ x=j$$
and
$$C_n^j(U_2^m) |xrangle|yrangle = 0 + |xrangle|yrangle = |xrangle|yrangle ~~textif~~ xneq j.$$

Gate $V_n^j(U_2^m)$ is basically the same as $C_n^j(U_2^m)$, though we consider first $m$ qubits as target and last $n$ qubits as control register in this case.

Now, if $j=11$ then $C_2^j(X)$ is exactly CCNOT gate on 3 qubits. Because we apply $X$ (i.e. negating the value) on the last qubit only if two first qubits are in $|11rangle$ state.