$\displaystyle\left[{{\underline{\alpha}}_{0}^{+}}\widetilde{\underline{A}}^{-}_{m},{{\underline{\alpha}}_{0}^{+}}\widetilde{\underline{A}}^{-}_{n}\right]=(m-n){{\underline{\alpha}}_{0}^{+}}\widetilde{\underline{A}}^{-}_{m+n}+\frac{26-D}{12}m^{3}\delta_{m+n,0}.$
(B.16)

##  Appendix C Derivation of hermiticity properties

We calculate the Hermitian conjugation of the $\widetilde{A}^{-}_{m}$
operators using explicitly, the mode expansion of the string solution.
Thereby, we note that,

$\displaystyle\left[e^{i\frac{m{\underline{L}}^{+}(z)}{\alpha^{\prime}p_{0}^{+}}}\right]^{\dagger}$
$\displaystyle=e^{\left[i\frac{m{\underline{L}}^{+}(z)}{\alpha^{\prime}p_{0}^{+}}\right]^{\dagger}}=\exp\left[\frac{im}{\alpha^{\prime}p_{0}^{+}}\left(\frac{1}{2}x^{+}_{0}-i\alpha^{\prime}p^{+}_{0}\ln(z)+i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{n\neq
0}\frac{\alpha^{+}_{n}}{n}z^{-n}\right)\right]^{\dagger}$
$\displaystyle=\exp\left[-\frac{im}{\alpha^{\prime}p_{0}^{+}}\left(\frac{1}{2}x^{+}_{0}+i\alpha^{\prime}p^{+}_{0}\ln({\bar{z}})-i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{n\neq
0}\frac{\alpha^{+}_{-n}}{n}{\bar{z}}^{-n}\right)\right]$
$\displaystyle=\exp\left[-\frac{im}{\alpha^{\prime}p_{0}^{+}}\left(\frac{1}{2}x^{+}_{0}-i\alpha^{\prime}p^{+}_{0}\ln(\frac{1}{{\bar{z}}})+i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{m\neq
0}\frac{\alpha^{+}_{m}}{m}\left(\frac{1}{{\bar{z}}}\right)^{-m}\right)\right]$
$\displaystyle\implies$
$\displaystyle\left[e^{i\frac{m{\underline{L}}^{+}(z)}{\alpha^{\prime}p_{0}^{+}}}\right]^{\dagger}=e^{-\frac{im{\underline{L}}^{+}\left(\frac{1}{{\bar{z}}}\right)}{\alpha^{\prime}p_{0}^{+}}},$
(C.1)

where we have used that $x_{0}^{\mu}$ and $p_{0}^{\mu}$ are Hermitian and
$(\alpha^{+}_{n})^{\dagger}=\alpha^{+}_{-n}$. We also compute,

$\displaystyle[\partial{\underline{L}}^{\mu}(z)]^{\dagger}$
$\displaystyle=\left[-i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{n\in\mathbb{Z}}\alpha^{\mu}_{n}z^{-n-1}\right]^{\dagger}=i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{n\in\mathbb{Z}}\alpha^{\mu}_{-n}{\bar{z}}^{-n-1}$
$\displaystyle=i\sqrt{\frac{\alpha^{\prime}}{2}}\frac{1}{{\bar{z}}^{2}}\sum_{m\in\mathbb{Z}}\alpha^{\mu}_{m}\left(\frac{1}{{\bar{z}}}\right)^{-m-1}=-\frac{1}{{\bar{z}}^{2}}\left[-i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{m\in\mathbb{Z}}\alpha^{\mu}_{m}\left(\frac{1}{{\bar{z}}}\right)^{-m-1}\right]$
$\displaystyle\implies$
$\displaystyle[\partial{\underline{L}}^{\mu}(z)]^{\dagger}=-\frac{1}{{\bar{z}}^{2}}\partial{\underline{L}}^{\mu}\left(\frac{1}{{\bar{z}}}\right),$
(C.2)

and the Hermitian conjugate of the second derivative,

$\displaystyle[\partial^{2}{\underline{L}}^{\mu}(z)]^{\dagger}$
$\displaystyle=-i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{n}(n+1)\alpha^{\mu}_{-n}{\bar{z}}^{-n-2}$
$\displaystyle=$
$\displaystyle-i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{m}(1-m)\alpha^{\mu}_{m}\left(\frac{1}{{\bar{z}}}\right)^{-m-2}\frac{1}{{\bar{z}}^{4}}$
$\displaystyle=\frac{1}{{\bar{z}}^{4}}i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{m}(m+1)\alpha^{\mu}_{m}\left(\frac{1}{{\bar{z}}}\right)^{-m-2}-2i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{m}\alpha^{\mu}_{m}\left(\frac{1}{{\bar{z}}}\right)^{-m-1}\frac{1}{{\bar{z}}^{3}}$
$\displaystyle\implies$
$\displaystyle[\partial^{2}{\underline{L}}^{\mu}(z)]^{\dagger}=\frac{1}{{\bar{z}}^{4}}\partial^{2}{\underline{L}}^{\mu}\left(\frac{1}{{\bar{z}}}\right)+2\frac{1}{{\bar{z}}^{3}}\partial{\underline{L}}^{\mu}\left(\frac{1}{{\bar{z}}}\right).$
(C.3)

Finally we compute the last piece,

$\displaystyle\left[\hat{Q}_{l;m}(E)\right]^{\dagger}=-\oint\frac{d{\bar{z}}}{2\pi
i}\frac{1}{{\bar{z}}^{l+1}}e^{-\frac{im{\underline{L}}^{+}\left(\frac{1}{{\bar{z}}}\right)}{\alpha^{\prime}p_{0}^{+}}},$
(C.4)

where the ${}^{\prime}-^{\prime}$ outside fixes the direction of the loop to
anti-clockwise after taking the complex conjugate. Then,

$\displaystyle\left[\hat{Q}_{l;m}(E)\right]^{\dagger}$
$\displaystyle=({\bar{z}})^{2}\oint\frac{d\left(\frac{1}{{\bar{z}}}\right)}{2\pi
i}\frac{1}{\left(\frac{1}{{\bar{z}}}\right)^{-l+1}\left(\frac{1}{{\bar{z}}}\right)^{-2}}e^{-\frac{im{\underline{L}}^{+}\left(\frac{1}{{\bar{z}}}\right)}{\alpha^{\prime}p_{0}^{+}}}$
$\displaystyle=\oint\frac{d\left(\frac{1}{{\bar{z}}}\right)}{2\pi
i}\frac{1}{\left(\frac{1}{{\bar{z}}}\right)^{-l+1}}e^{-\frac{im{\underline{L}}^{+}\left(\frac{1}{{\bar{z}}}\right)}{\alpha^{\prime}p_{0}^{+}}}$
$\displaystyle\implies$
$\displaystyle\left[\hat{Q}_{l;m}(E)\right]^{\dagger}=\hat{Q}_{-l;-m}(E).$
(C.5)

To compute the Hermitian conjugate of $\widehat{\underline{A}}^{-}_{m}$, we
first note that the action of H.C on a normal ordered product of two operators
$A:=\partial{\underline{L}}^{-}$ and
$B:=e^{\frac{im{\underline{L}}^{+}}{\alpha^{\prime}p_{0}^{+}}}$ is given by,

$\displaystyle(:AB:)^{\dagger}=B^{\dagger}A^{\dagger}-\langle{AB}\rangle^{\dagger}$
$\displaystyle=\leavevmode\nobreak\
:A^{\dagger}B^{\dagger}:+[B^{\dagger},A^{\dagger}].$ (C.6)

Using the above results, we then observe that,

$\displaystyle:\left(\partial{\underline{L}}^{-}(z)e^{\frac{im{\underline{L}}^{+}(z)}{\alpha^{\prime}p_{0}^{+}}}\right):\leavevmode\nobreak\
$ $\displaystyle=\leavevmode\nobreak\
:\left(-\frac{1}{{\bar{z}}^{2}}\right)\partial{\underline{L}}^{-}\left(\frac{1}{{\bar{z}}}\right)e^{-\frac{im{\underline{L}}^{+}\left(\frac{1}{{\bar{z}}}\right)}{\alpha^{\prime}p_{0}^{+}}}:+\frac{im}{\alpha^{\prime}p_{0}^{+}}\left(-\frac{1}{{\bar{z}}^{2}}\right)(-i\alpha^{\prime})\frac{{\bar{z}}}{2}ie^{-\frac{im{\underline{L}}^{+}\left(\frac{1}{{\bar{z}}}\right)}{\alpha^{\prime}p_{0}^{+}}}$
$\displaystyle\implies$
$\displaystyle[\widehat{\underline{A}}^{-}_{m}]^{\dagger}=\widehat{\underline{A}}^{-}_{-m}+\frac{m}{\alpha^{+}_{0}}\widehat{Q}_{0;-m}$
(C.7)

Finally, we can see that,

$\displaystyle[{\underline{A}}^{-}_{m}(E)]^{\dagger}=$
$\displaystyle[\widehat{\underline{A}}^{-}_{m}(E)]^{\dagger}+i\sqrt{\frac{2}{\alpha^{\prime}}}\oint\frac{d{\bar{z}}}{2\pi
i}:\left[\frac{im}{4p_{0}^{+}}\left(-\frac{1}{{\bar{z}}^{2}}\frac{\partial^{2}{\underline{L}}^{+}}{\partial{\underline{L}}^{+}}-\frac{2}{{\bar{z}}}\right)\right]e^{-\frac{im{\underline{L}}^{+}\left(\frac{1}{{\bar{z}}}\right)}{\alpha^{\prime}p_{0}^{+}}}:$
$\displaystyle=\widehat{\underline{A}}^{-}_{-m}+$
$\displaystyle\frac{m}{\alpha^{+}_{0}}\widehat{Q}_{0;-m}+i\sqrt{\frac{2}{\alpha^{\prime}}}\oint\frac{d\left(\frac{1}{{\bar{z}}}\right)}{2\pi
i}:\left(-\frac{i(-m)}{4p_{0}^{+}}\frac{\partial^{2}{\underline{L}}^{+}}{\partial{\underline{L}}^{+}}\right)e^{-\frac{im{\underline{L}}^{+}\left(\frac{1}{{\bar{z}}}\right)}{\alpha^{\prime}p_{0}^{+}}}:$
$\displaystyle+i\sqrt{\frac{2}{\alpha^{\prime}}}\oint\frac{d\left(\frac{1}{{\bar{z}}}\right)}{2\pi
i}:\frac{2im}{4p_{0}^{+}}\frac{1}{\left(\frac{1}{{\bar{z}}}\right)^{1}}e^{-\frac{im{\underline{L}}^{+}\left(\frac{1}{{\bar{z}}}\right)}{\alpha^{\prime}p_{0}^{+}}}:$
$\displaystyle=$
$\displaystyle{\underline{A}}^{-}_{-m}(E)+\frac{m}{\alpha^{+}_{0}}\widehat{Q}_{0;-m}-\frac{m}{\alpha^{+}_{0}}\widehat{Q}_{0;-m}$
(C.8) $\displaystyle\implies$
$\displaystyle[{\underline{A}}^{-}_{m}(E)]^{\dagger}={\underline{A}}^{-}_{-m}(E)$
(C.9) $\displaystyle\implies$
$\displaystyle[{\underline{A}}^{-}_{m}(E)]^{{\dagger}{\dagger}}={\underline{A}}^{-}_{m}(E).$
(C.10)

We also note that the $\widetilde{\underline{A}}^{-}_{m}$ follows the same
algebra under Hermitian conjugation as ${\underline{A}}^{-}_{m}$. This is
evident from its definition and the properties of the involved quantities
derived in the preceding sections of the Appendix.

##  Appendix D Details on Lorentz algebra

###  D.1 $M^{+2}$ Infinitesimal rotation

To better understand the discussion above, we explicitly compute the variation
of the flat space DDF operators using an infinitesimal rotation such that,

$E^{1^{\prime}}=E^{1}+\sqrt{2}\epsilon E^{2},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
E^{2^{\prime}}=E^{2}-\sqrt{2}\epsilon E^{1},$ (D.1)

where the factor $\sqrt{2}$ has been chosen so that,

$E^{\pm^{\prime}}=E^{\pm}\pm\epsilon E^{2},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
E^{2^{\prime}}=E^{2}-\epsilon E^{+}+\epsilon E^{-}.$ (D.2)

Under this condition we can write,

$\displaystyle e^{im\frac{E^{+^{\prime}}\cdot
L(z)}{\alpha^{\prime}E^{+^{\prime}}\cdot p_{0}}}=$ $\displaystyle
e^{im\frac{E^{+}\cdot L(z)}{\alpha^{\prime}E^{+}\cdot
p_{0}}}\left[1+\epsilon\frac{im}{\alpha^{\prime}({\underline{p}}^{+}_{0})^{2}}\left({\underline{p}}_{0}^{+}{\underline{L}}^{2}-{\underline{p}}_{0}^{2}{\underline{L}}^{+}\right)+O(\epsilon^{2})\right]$
$\displaystyle=e^{im\frac{E^{+}\cdot L(z)}{\alpha^{\prime}E^{+}\cdot p_{0}}}$
$\displaystyle\left[1+\epsilon\frac{im}{\alpha^{\prime}({\underline{p}}^{+}_{0})^{2}}\left({\underline{p}}_{0}^{+}{\underline{L}}_{(\neq)}^{2}-{\underline{p}}_{0}^{2}{\underline{L}}_{(\neq)}^{+}\right)+O(\epsilon^{2})\right],$
(D.3)

where we have defined,

${\underline{L}}_{(\neq)}(z)={\underline{L}}(z)-i\alpha^{\prime}{\underline{p}}_{0}\ln(z),$
(D.4)

i.e. the part of ${\underline{L}}$ without the logarithmic cut. Note the
normal ordering relations,

$\displaystyle{\underline{L}}_{(\neq)}(z)\,{\underline{L}}_{(\neq)}(w)=$
$\displaystyle:{\underline{L}}_{(\neq)}(z)\,{\underline{L}}_{(\neq)}(w):-\frac{\alpha^{\prime}}{2}\ln\left(1-\frac{z}{w}\right),\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ |z|>|w|$
$\displaystyle{\underline{L}}(z)\,{\underline{L}}_{(\neq)}(w)=$
$\displaystyle:{\underline{L}}(z)\,{\underline{L}}_{(\neq)}(w):-\frac{\alpha^{\prime}}{2}\ln\left(z-w\right),$
(D.5)

where the first one has no cuts.

It is important to use ${\underline{L}}_{(\neq)}$ if we want to use the usual
formula

$\left[\oint_{z=0}{\cal A}(z),\oint_{w=0}{\cal
E}(w)\right]=\oint_{w=0}\oint_{z=w}R[{\cal A}(z){\cal E}(w)],$ (D.6)

since its derivation requires the absence of cuts.

From the previous expression we get,

$\displaystyle\delta_{\epsilon}A^{i}_{m}(E)=$
$\displaystyle\leavevmode\nobreak\ A^{i}_{m}(E^{\prime})-A^{i}_{m}(E)$
$\displaystyle=\epsilon$
$\displaystyle\frac{im}{\alpha^{\prime}({\underline{p}}^{+}_{0})^{2}}i\sqrt{\frac{2}{\alpha^{\prime}}}\oint_{z=0}\frac{dz}{2\pi
i}:\partial_{z}{\underline{L}}^{i}(z)\left({\underline{p}}_{0}^{+}{\underline{L}}_{(\neq)}^{2}(z)-{\underline{p}}_{0}^{2}{\underline{L}}_{(\neq)}^{+}(z)\right)e^{im\frac{{\underline{L}}^{+}(z)}{\alpha^{\prime}{\underline{p}}^{+}_{0}}}:$
$\displaystyle+\delta^{i2}(-\epsilon)i\sqrt{\frac{2}{\alpha^{\prime}}}\oint_{z=0}\frac{dz}{2\pi
i}:\left(\partial_{z}{\underline{L}}^{+}(z)-\partial_{z}{\underline{L}}^{-}(z)\right)e^{im\frac{{\underline{L}}^{+}(z)}{\alpha^{\prime}{\underline{p}}^{+}_{0}}}:,$
(D.7)

where the contribution in the last line is only for $i=2$ and does not use
${\underline{L}}_{(\neq)}$, but ${\underline{L}}$ since it comes from the
rotation of $\partial{\underline{L}}^{2}$. We shall study the variation above
separately for two cases ($i\neq 2$ and $i=2$).

####  D.1.1 Rewriting the variation $\delta_{\epsilon}A^{i}_{m}(E)$ for $i\neq
2$

If the mapping $A^{i}_{n}\leftrightarrow\alpha^{i}_{n({lightcone})}$ is indeed
true, we should have that the previous variation
$\delta_{\epsilon}A^{i}_{m}(E)=\delta_{\epsilon}\alpha^{i}_{m({lightcone})}$,
up to terms which vanish in the lightcone gauge.

To check this, we evaluate the commutator,

$[A^{i}_{m}(E),\delta_{\epsilon}A^{j}_{n}(E)],$ (D.8)

and then determine the coefficients $c_{l}$ and ${\cal N}$ in

$\delta_{\epsilon}{\underline{A}}^{j}_{n}(E)=\sum_{l}c_{l}\,{\underline{A}}^{j}_{l}(E)\,{\underline{A}}^{2}_{n-l}(E)+{\cal
N}{\tilde{\underline{A}}}^{-}_{n}(E),$ (D.9)

in order to reproduce the previous commutators. The expression in (D.9) is an
ansatz based on the operator’s level and $SO(D-2)$ rotations.

We evaluate (D.7) for $j\neq 2$ as follows,

$\displaystyle[$ $\displaystyle{\underline{A}}^{i}_{m}(E),\leavevmode\nobreak\
\delta_{\epsilon}{\underline{A}}^{j}_{n}(E)]=\epsilon\frac{in}{\alpha^{\prime}({\underline{p}}^{+}_{0})^{2}}\cdot\oint_{w=0}\frac{dw}{2\pi
i}\oint_{z=w}\frac{dz}{2\pi
i}\Biggl{[}\frac{\delta^{ij}}{(z-w)^{2}}\left({\underline{p}}_{0}^{+}{\underline{L}}_{(\neq)}^{2}(w)-{\underline{p}}_{0}^{2}{\underline{L}}_{(\neq)}^{+}(w)\right)$
$\displaystyle+\frac{\delta^{i2}}{z-w}{\underline{p}}_{0}^{+}\partial{\underline{L}}^{j}(w)+\frac{-2}{\alpha^{\prime}}:\partial{\underline{L}}^{i}(z)\partial{\underline{L}}^{j}(w)\left({\underline{p}}_{0}^{+}{\underline{L}}_{(\neq)}^{2}(w)-{\underline{p}}_{0}^{2}{\underline{L}}_{(\neq)}^{+}(w)\right):\Biggr{]}\,e^{im\frac{{\underline{L}}^{+}(z)}{\alpha^{\prime}p^{+}_{0}}}e^{in\frac{{\underline{L}}^{+}(w)}{\alpha^{\prime}{\underline{p}}^{+}_{0}}}$
$\displaystyle\leavevmode\nobreak\
=\epsilon\frac{in}{\alpha^{\prime}({\underline{p}}^{+}_{0})^{2}}\oint_{w=0}\frac{dw}{2\pi
i}\Biggl{[}\delta^{ij}\,\left({\underline{p}}_{0}^{+}{\underline{L}}_{(\neq)}^{2}(w)-{\underline{p}}_{0}^{2}{\underline{L}}_{(\neq)}^{+}(w)\right)\,\frac{im}{\alpha^{\prime}{\underline{p}}^{+}_{0}}\,\partial{\underline{L}}^{+}(w)\,e^{i(m+n)\frac{{\underline{L}}^{+}(w)}{\alpha^{\prime}{\underline{p}}^{+}_{0}}}$
$\displaystyle\phantom{\epsilon\frac{in}{\alpha^{\prime}({\underline{p}}^{+}_{0})^{2}}}+\delta^{i2}\,{\underline{p}}_{0}^{+}\,\partial{\underline{L}}^{j}(w)\,e^{i(m+n)\frac{{\underline{L}}^{+}(w)}{\alpha^{\prime}{\underline{p}}^{+}_{0}}}\Biggr{]}.$
(D.10)

There are now two cases: $m+n\neq 0$ and $m+n=0$. For the first case, we can
integrate by parts since,
$\left({\underline{p}}_{0}^{+}{\underline{L}}_{(\neq)}^{2}(w)-{\underline{p}}_{0}^{2}{\underline{L}}_{(\neq)}^{+}(w)\right)$
has no cut and we can write,

$\displaystyle[{\underline{A}}^{i}_{m},\leavevmode\nobreak\
\delta_{\epsilon}{\underline{A}}^{j}_{n}]=\epsilon\frac{in}{\sqrt{2\alpha^{\prime}}({\underline{p}}^{+}_{0})^{2}}\Biggl{[}\left(-{\underline{p}}_{0}^{+}{\underline{A}}^{2}_{m+n}\frac{m}{m+n}\delta^{ij}+{\underline{p}}_{0}^{+}{\underline{A}}^{j}_{m+n}\delta^{i2}\right)+{\underline{p}}_{0}^{2}{\underline{A}}^{+}_{m+n}\frac{m}{m+n}\delta^{ij}\Biggr{]},$
(D.11)

where we have defined,

${\underline{A}}^{+}_{m}(E)=i\sqrt{\frac{2}{\alpha^{\prime}}}\oint\frac{dz}{2\pi
i}\partial{\underline{L}}^{+}(z)e^{im\frac{{\underline{L}}^{+}(z)}{\alpha^{\prime}{\underline{p}}^{+}_{0}}}=\delta_{m,0}{\underline{\alpha}}^{+}_{0},$
(D.12)

which is trivial for $m\neq 0$.

In order to determine the $c_{l}$ coefficients in eq. (D.9), we insert the
ansatz into the commutator and obtain,

$\displaystyle[A^{i}_{m}(E),$
$\displaystyle\delta_{\epsilon}A^{j}_{n}(E)]=m\delta^{ij}\,c_{-m}\,{\underline{A}}^{2}_{m+n}(E)+m\delta^{i2}\,c_{n+m}\,{\underline{A}}^{j}_{m+n}(E),$
$\displaystyle\implies$ $\displaystyle
m\,c_{-m}=-\epsilon\frac{i}{\alpha^{\prime}{\underline{p}}^{+}_{0}}\frac{mn}{m+n}\left(-i\sqrt{\frac{\alpha^{\prime}}{2}}\right);\leavevmode\nobreak\
m\,c_{n+m}=+\epsilon\frac{i}{\alpha^{\prime}{\underline{p}}^{+}_{0}}n\left(-i\sqrt{\frac{\alpha^{\prime}}{2}}\right),$
(D.13)

when compared with the actual result. Taking the ratio we get

$\frac{c_{-m}}{c_{n+m}}=\frac{-m}{n+m},$ (D.14)

which suggests two possible classes of solutions and a combination of the two.
The two basic possible solutions are,

$c_{l}=\tilde{c}_{1}\,l,\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ c_{l}=\hat{c}_{1}\frac{1}{n-l}.$
(D.15)

Inserting them into the original equations we get,

$\displaystyle c_{l}=\tilde{c}_{1}\,l$
$\displaystyle\Rightarrow\tilde{c}_{1}=\epsilon\frac{1}{\sqrt{2\alpha^{\prime}}{\underline{p}}_{0}^{+}}\frac{n}{m}\frac{1}{m+n},$
$\displaystyle c_{l}=\hat{c}_{1}\frac{1}{n-l}$
$\displaystyle\Rightarrow\hat{c}_{1}=-\epsilon\frac{1}{\sqrt{2\alpha^{\prime}}{\underline{p}}_{0}^{+}}n.$
(D.16)

Since $c_{l}$ may depend on $n$, but must not depend on $m$, only the second
solution is viable.

Now, for the second case ($n+m=0$), we see that we get two contributions. The
first one corresponds to the term $l=0$ in (D.9). The second one is related to
the $\underline{M}^{+2}$ generator. Explicitly we get

$\displaystyle[{\underline{A}}^{i}_{-n}(E),\delta_{\epsilon}{\underline{A}}^{j}_{n}(E)]=$
$\displaystyle+\epsilon\frac{n}{\sqrt{2\alpha^{\prime}}{\underline{p}}_{0}^{+}}\delta^{i2}{\underline{A}}^{j}_{0}(E)$
$\displaystyle+$ $\displaystyle\leavevmode\nobreak\
\epsilon\frac{n^{2}}{(\alpha^{\prime}{\underline{p}}_{0}^{+})^{2}}\delta^{ij}\oint\frac{dz}{2\pi
i}({\underline{p}}_{0}^{+}{\underline{L}}^{2}_{(\neq)}(z)-{\underline{p}}_{0}^{2}{\underline{L}}^{+}_{(\neq)}(z))\partial{\underline{L}}^{+}(z).$
(D.17)

From the first term above, we see that,
$c_{0}=-\epsilon\frac{1}{\sqrt{2\alpha^{\prime}}{\underline{p}}^{+}_{0}}$ as
expected for the $l=0$ term.

We are left with the term
$\epsilon\frac{n^{2}}{\sqrt{2\alpha^{\prime}}{\underline{p}}_{0}^{+}}\delta^{ij}I^{(2)}$,
where,

$I^{(2)}=\oint\frac{dz}{2\pi
i}({\underline{p}}_{0}^{+}{\underline{L}}^{2}_{(\neq)}(z)-{\underline{p}}_{0}^{2}{\underline{L}}^{+}_{(\neq)}(z))\partial{\underline{L}}^{+}(z)\equiv\frac{1}{2}{\underline{p}}_{0}^{+}\underline{M}^{2+}(E),$
(D.18)

and $\underline{M}^{+2}(E)$ is the Lorentz generator in the frame $E$. To see
why this is so, we observe that,

$\displaystyle\oint\frac{dz}{2\pi
i}{\underline{L}}^{+}_{(\neq)}(z)\partial{\underline{L}}^{+}(z)=$
$\displaystyle\oint\frac{dz}{2\pi
i}{\underline{L}}^{+}_{(\neq)}(z)\partial\left({\underline{L}}_{(\neq)}^{+}(z)-i\alpha^{\prime}{\underline{p}}_{0}^{+}\ln(z)\right)$
$\displaystyle=$ $\displaystyle\leavevmode\nobreak\
-i\alpha^{\prime}{\underline{p}}_{0}^{+}\oint\frac{dz}{2\pi
iz}{\underline{L}}^{+}_{(\neq)}(z)=-\frac{1}{2}i\alpha^{\prime}{\underline{p}}_{0}^{+}{\underline{x}}_{0}^{+}.$
(D.19)

Then the first term can be rewritten as,

$\displaystyle\oint\frac{dz}{2\pi
i}{\underline{L}}^{2}_{(\neq)}(z)\partial{\underline{L}}^{+}(z)=+\oint\frac{dz}{2\pi
i}{\underline{L}}^{2}_{(\neq)}(z)\partial{\underline{L}}_{(\neq)}^{+}(z)-\frac{1}{2}i\alpha^{\prime}{\underline{p}}_{0}^{+}{\underline{x}}_{0}^{2}$
$\displaystyle=-\oint\frac{dz}{2\pi
i}{\underline{L}}^{+}_{(\neq)}(z)\partial{\underline{L}}_{(\neq)}^{2}(z)-\frac{1}{2}i\alpha^{\prime}{\underline{p}}_{0}^{+}{\underline{x}}_{0}^{2},$
(D.20)

or taking the semi-sum as,

$\oint\frac{dz}{2\pi
i}{\underline{L}}^{2}_{(\neq)}(z)\partial{\underline{L}}^{+}(z)=\frac{1}{2}\oint\frac{dz}{2\pi
i}({\underline{L}}^{2}_{(\neq)}(z)\partial{\underline{L}}_{(\neq)}^{+}(z)-{\underline{L}}^{+}_{(\neq)}(z)\partial{\underline{L}}_{(\neq)}^{2}(z))-\frac{1}{2}i\alpha^{\prime}{\underline{p}}_{0}^{+}{\underline{x}}_{0}^{2}.$
(D.21)

Assembling all the contributions and using $\partial{\underline{L}}$ instead
of $\partial{\underline{L}}_{(\neq)}$ we get,

$\displaystyle I^{(2)}=\frac{1}{2}{\underline{p}}_{0}^{+}$
$\displaystyle\left[\oint\frac{dz}{2\pi
i}({\underline{L}}^{2}_{(\neq)}(z)\partial{\underline{L}}^{+}(z)-{\underline{L}}^{+}_{(\neq)}(z)\partial{\underline{L}}^{2}(z))+\frac{1}{2}i\alpha^{\prime}({\underline{p}}_{0}^{2}{\underline{x}}_{0}^{+}-{\underline{p}}_{0}^{+}{\underline{x}}_{0}^{2})\right]$
$\displaystyle=\frac{1}{2}{\underline{p}}_{0}^{+}\underline{M}^{2+}(E),$
(D.22)

which matches (LABEL:eq:Lorentz_generator) up to an overall factor in the
frame $E$.

Noting that $M^{+2}$ rotates $x^{2}\rightarrow x^{+}$ and $x^{-}\rightarrow
x^{2}$ but keeps $x^{+}$ fixed, the term $I^{(2)}$ commutes with all DDFs
$A^{i}_{m}$ with $i\neq 2$. Explicitly we get,

$[{\underline{A}}^{i}_{-m}(E),I^{(2)}]=-\frac{1}{2}\sqrt{\frac{\alpha^{\prime}}{2}}\delta^{i2}({\underline{\alpha}}^{+}_{0})^{2}\delta_{m,0}.$
(D.23)

Assembling all contributions for $j\neq 2$ we get,

$\displaystyle\delta_{\epsilon}{\underline{A}}^{j}_{n}(E)$
$\displaystyle=-\epsilon\frac{n}{\sqrt{2\alpha^{\prime}}{\underline{p}}^{+}_{0}}\left[\sum_{l\neq
n}\frac{1}{n-l}{\underline{A}}^{j}_{l}(E)\,{\underline{A}}^{2}_{n-l}(E)+\frac{1}{\sqrt{2\alpha^{\prime}}{\underline{p}}^{+}_{0}}\underline{M}^{+2}(E){\underline{A}}^{j}_{n}(E)\right],$
(D.24)

where the contribution from ${\underline{A}}^{-}_{n}$ is absent since in
$\delta_{\epsilon}{\underline{A}}^{j\neq 2}_{n}(E)$ there are no
${\underline{\alpha}}^{-}$. There are also no normal ordering issues in (D.24)
since, $\underline{M}^{+2}(E)$ commutes with all DDFs as noted above.

####  D.1.2 Normal ordering subtlety in the $i=2$ case

In (D.7) we would like to consider the details of the variation
$\delta_{\epsilon}A^{2}_{m}(E)$ since some normal ordering is required and
because the integrand
$\partial{\underline{L}}^{-}e^{ik^{.}{\underline{L}}^{+}}$ is not a good
conformal operator - it has a cubic pole in the OPE with the stress energy
tensor.

The key observation is that, $\lim_{z\rightarrow
w}\partial{\underline{L}}^{2}(z)e^{ik^{.}{\underline{L}}^{+}(w)}$ is a
perfectly well defined conformal operator and therefore its infinitesimal
Lorentz rotation must be as well. We expect it to be given by the normal
ordering of the single components rotations but other quantum i.e. vanishing
classically terms may in principle appear.

We start by computing the infinitesimal Lorentz rotation,

$\displaystyle[\sqrt{2}\epsilon M^{12}$
$\displaystyle,\partial{\underline{L}}^{2}(z)e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}(w)}]=\delta_{\epsilon}(\partial{\underline{L}}^{2}(z))e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}(w)}+\partial{\underline{L}}^{2}(z)\delta_{\epsilon}\left(e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}(w)}\right)$
$\displaystyle=$
$\displaystyle-\epsilon(\partial{\underline{L}}^{+}(z)-\partial{\underline{L}}^{-}(z))\,e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}(w)}$
$\displaystyle+\partial{\underline{L}}^{2}(z)\,\left(\epsilon\frac{im}{\alpha^{\prime}({\underline{p}}^{+}_{0})^{2}}\left({\underline{p}}_{0}^{+}{\underline{L}}_{(\neq)}^{2}(w)-{\underline{p}}_{0}^{2}{\underline{L}}_{(\neq)}^{+}(w)\right)e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}(w)}\right).$
(D.25)

We can then evaluate the singular part of the OPE to get,

$\displaystyle\left.\left[\sqrt{2}\epsilon
M^{12},\partial{\underline{L}}^{2}(z)e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}(w)}\right]\right|_{\mbox{sing}}=\frac{-\epsilon\alpha^{\prime}}{2(z-w)}\left(i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}\eta^{+-}+\frac{im}{\alpha^{\prime}({\underline{p}}^{+}_{0})^{2}}{\underline{p}}_{0}^{+}\right)e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}(w)},$
(D.26)

which vanishes and therefore the infinitesimal Lorentz rotation is given by
the normal ordering of the rotations of the single components.

Finally the infinitesimal Lorentz rotation is,

$\displaystyle\delta_{\epsilon}\left(\partial{\underline{L}}^{2}e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}}\right)=$
$\displaystyle\epsilon\,:\left[\frac{im}{\alpha^{\prime}({\underline{p}}^{+}_{0})^{2}}\partial{\underline{L}}^{2}\left({\underline{p}}_{0}^{+}{\underline{L}}_{(\neq)}^{2}-{\underline{p}}_{0}^{2}{\underline{L}}_{(\neq)}^{+}\right)-(\partial{\underline{L}}^{+}-\partial{\underline{L}}^{-})\right]e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}}:$
$\displaystyle=$
$\displaystyle\epsilon\,:\left[\partial{\underline{L}}^{-}+\frac{im}{\alpha^{\prime}({\underline{p}}^{+}_{0})^{2}}({\underline{p}}_{0}^{+}{\underline{L}}_{(\neq)}^{2}-{\underline{p}}_{0}^{2}{\underline{L}}_{(\neq)}^{+})\partial{\underline{L}}^{2}\right]e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}}:$
$\displaystyle-\epsilon\partial\left(\frac{\alpha^{\prime}{\underline{p}}_{0}^{+}}{im}e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}}\right),$
(D.27)

which contains pieces which are not good conformal operators. However the sum
of first two operators is free from cubic poles,

$T=-\frac{2}{\alpha^{\prime}}:(\partial L)^{2}:$ to be

$\displaystyle-\frac{\alpha^{\prime}}{2}T(z)\,$
$\displaystyle\left(\partial{\underline{L}}^{-}e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}}\right)(w)=\left.2e^{\delta\cdot\partial
L(z)}\right|_{\delta_{\mu}\delta_{\nu}\rightarrow\eta_{\mu\nu}}\,\left.e^{\epsilon\cdot\partial
L+i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}}\right|_{\epsilon_{-}}$
$\displaystyle=$ $\displaystyle
2\left.:e^{\delta\cdot{\underline{L}}(z)+\epsilon\cdot\partial
L(w)+i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}(w)}:e^{-\frac{\alpha^{\prime}}{2}\frac{\epsilon\cdot\delta}{(z-w)^{2}}-\frac{\alpha^{\prime}}{2}\frac{i\delta\cdot
k}{z-w}}\right|_{\delta_{\mu}\delta_{\nu}\rightarrow\eta_{\mu\nu},\leavevmode\nobreak\
\epsilon_{-},k_{+}\rightarrow\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}}$
$\displaystyle=$ $\displaystyle
2\left(\frac{\alpha^{\prime}}{2}\right)^{2}\frac{i\epsilon\cdot
k}{(z-w)^{3}}+\dots$ $\displaystyle=$ $\displaystyle
2\left(\frac{\alpha^{\prime}}{2}\right)^{2}\frac{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}\eta^{+-}}{(z-w)^{3}}+\dots.$
(D.28)

In a similar way we get

$\displaystyle-\frac{\alpha^{\prime}}{2}T(z)\,$
$\displaystyle\left(\partial{\underline{L}}^{2}L^{2}_{(\neq)}e^{i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}}\right)(w)=\left.2e^{\delta\cdot\partial
L(z)}\right|_{\delta_{\mu}\delta_{\nu}\rightarrow\eta_{\mu\nu}}\,\left.e^{\epsilon_{(1)}\cdot\partial
L+\epsilon_{(2)}\cdot\partial
L_{(\neq)}+i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}{\underline{L}}^{+}}\right|_{\epsilon_{(1)2},\epsilon_{(2)2}}$
$\displaystyle=$ $\displaystyle
2\left(\frac{\alpha^{\prime}}{2}\right)^{2}\frac{\epsilon_{(1)}\cdot\epsilon_{(2)}}{(z-w)^{3}}+\dots$
$\displaystyle=$ $\displaystyle
2\left(\frac{\alpha^{\prime}}{2}\right)^{2}\frac{1}{(z-w)^{3}}+\dots.$ (D.29)

Now we can examine the cubic pole contributions and see that,

$\displaystyle-\frac{\alpha^{\prime}}{2}T(z)\,$
$\displaystyle\left(\dots\right)=2\left(\frac{\alpha^{\prime}}{2}\right)^{2}\frac{-i\frac{m}{\alpha^{\prime}{\underline{p}}_{0}^{+}}}{(z-w)^{3}}+\frac{im}{\alpha^{\prime}({\underline{p}}^{+}_{0})^{2}}\,{\underline{p}}_{0}^{+}\times
2\left(\frac{\alpha^{\prime}}{2}\right)^{2}\frac{1}{(z-w)^{3}}+\dots,$ (D.30)

which in fact vanishes.

####  D.1.3 Rewriting the variation $\delta_{\epsilon}A^{j}_{m}(E)$ for $j=2$

Finally we consider the $j=2$ case where there is a further contribution in
eq. (D.7), when $i=2$. If we proceed as before, we can evaluate the full
commutator as,

$\displaystyle[A^{i}_{m}(E),\delta_{\epsilon}A^{j=2}_{n}]=$
$\displaystyle\epsilon\frac{in}{{\underline{\alpha}}^{+}_{0}}\left(-{\underline{A}}^{2}_{m+n}\frac{m}{m+n}+{\underline{A}}^{2}_{m+n}\right)\delta^{i2}-m\delta^{j2}\frac{\epsilon}{{\underline{\alpha}}^{+}_{0}}{\underline{A}}^{i}_{m+n}.$
(D.31)

We observe that the ansatz in (D.9) can generate the first term (in
parentheses), wherein we get explicitly,

$\displaystyle{c}_{-m}+{c}_{m+n}=-\frac{i\epsilon}{{\underline{\alpha}}^{+}_{0}}\frac{n^{2}}{m(m+n)}.$
(D.32)

Once again, there are two basic possible linear homogeneous solutions to this
recurrence relation, given by (D.15). Furthermore, we must choose the second
possibility for the exact same reason as before, to obtain,

$\displaystyle
c_{l}=-\frac{i\epsilon}{{\underline{\alpha}}^{+}_{0}}\frac{n}{n-l}.$ (D.33)

We see that this term is neatly absorbed in the expansion of
$\delta_{\epsilon}A^{j}$, which now reads,

$\displaystyle\delta_{\epsilon}{\underline{A}}^{j}_{n}(E)$
$\displaystyle=-i\epsilon\frac{n}{\sqrt{2\alpha^{\prime}}{\underline{p}}^{+}_{0}}\left[\sum_{l\neq
n}\frac{1}{n-l}{\underline{A}}^{j}_{l}(E)\,{\underline{A}}^{2}_{n-l}(E)+\frac{1}{\sqrt{2\alpha^{\prime}}{\underline{p}}^{+}_{0}}\underline{M}^{+2}(E){\underline{A}}^{j}_{n}(E)\right],$
(D.34)

and is no longer limited to $j\neq 2$.

Finally, we consider the last piece in (D.31) - since there are no more terms
like $\delta^{i\dots}$, we assume an expression like,

$\delta_{\epsilon}{\underline{A}}^{j=2}_{n}(E)|_{\mbox{extra}}=\delta^{j2}\left[\sum_{l}d_{l}\,{\underline{A}}^{k}_{l}(E)\,{\underline{A}}^{k}_{n-l}(E)+\bar{\cal
N}{\tilde{\underline{A}}}^{-}_{n}(E)\right],$ (D.35)

where, ’extra’ labels the last term of (D.31). Comparison with the true
expression gives,

$d_{m+n}+d_{-m}=\frac{-i\epsilon}{\sqrt{2\alpha^{\prime}}{\underline{p}}_{0}^{+}},$
(D.36)

whose simplest solution is,

$d_{l}=\frac{1}{2}\frac{-i\epsilon}{\sqrt{2\alpha^{\prime}}{\underline{p}}_{0}^{+}}.$
(D.37)

Putting everything together, the complete variation is given by,

$\displaystyle\delta_{\epsilon}{\underline{A}}^{j}_{n}(E)=-i\epsilon\frac{n}{\sqrt{2\alpha^{\prime}}{\underline{p}}^{+}_{0}}$
$\displaystyle\left[\sum_{l\neq
n}\frac{1}{n-l}{\underline{A}}^{j}_{l}(E)\,{\underline{A}}^{2}_{n-l}(E)+\frac{1}{\sqrt{2\alpha^{\prime}}{\underline{p}}^{+}_{0}}\underline{M}^{+2}(E){\underline{A}}^{j}_{n}(E)\right]$
$\displaystyle-$
$\displaystyle\delta^{j2}\left[\frac{1}{2}\frac{i\epsilon}{\sqrt{2\alpha^{\prime}}{\underline{p}}_{0}^{+}}\sum_{l}{\underline{A}}^{k}_{l}(E)\,{\underline{A}}^{k}_{n-l}(E)+\bar{\cal
N}{\tilde{\underline{A}}}^{-}_{n}(E)\right].$ (D.38)

##  Appendix E Details on Lorentz generators expressed DDF and Brower
operators

In this appendix we report the details of how to write the action of Lorentz
generators on DDF and Brower operators using DDF and Brower operators. This is
not the same as expressing the Lorentz generators using DDF and Brower
operators since they span a subset of all covariant operators.

###  E.1 Preliminaries

The covariant Lorentz generator

$M^{\mu\nu}=x_{0}^{\mu}p_{0}^{\nu}-x_{0}^{\nu}p_{0}^{\nu}+i\sum_{n\neq
0}\frac{\alpha^{\mu}_{n}\alpha^{\nu}_{-n}}{n},$ (E.1)

satisfies

$\displaystyle[M^{\mu\nu},x_{0}^{\lambda}]$
$\displaystyle=2i\delta^{\lambda[\mu}x_{0}^{\nu]},$
$\displaystyle[M^{\mu\nu},\alpha_{m}^{\lambda}]$
$\displaystyle=2i\delta^{\lambda[\mu}\alpha_{m}^{\nu]},$ (E.2)

or for the chiral field

$\displaystyle[M^{\mu\nu},L(z)^{\lambda}]$
$\displaystyle=2i\delta^{\lambda[\mu}L(z)^{\nu]}.$ (E.3)

####  E.1.1 A comment on commutators involving the not derived $L(z)$

Naively one would expect that the operator

$\displaystyle{\widehat{\hat{M}}}^{\mu\nu}=$ $\displaystyle{\cal
N}\oint_{z=0}\frac{dz}{2\pi i}\left(L^{\mu}(z)\partial
L^{\nu}(z)-L^{\nu}(z)\partial L^{\mu}(z)\right),$ (E.4)

be the Lorentz generator. But it is not so because of the term
$-i\alpha^{\prime}p_{0}^{\mu}\,\ln z$ which implies a cut. Then one could
think of

$\displaystyle\hat{M}^{\mu\nu}$ $\displaystyle={\cal
N}\oint_{z=0}\frac{dz}{2\pi i}\left(L^{\mu}_{(\neq)}(z)\partial
L^{\nu}(z)-L^{\nu}_{(\neq)}(z)\partial L^{\mu}(z)\right),\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ {\cal N}=\frac{i}{\alpha^{\prime}}$
(E.5)

be the Lorentz generator. But it is not so. We can evaluate directly the
operator in terms of the oscillators as

$\displaystyle\frac{1}{{\cal N}}\hat{M}^{\mu\nu}=$
$\displaystyle-i\frac{\alpha^{\prime}}{2}\left(x_{0}^{\mu}p_{0}^{\nu}-x_{0}^{\nu}p_{0}^{\mu}\right)+\alpha^{\prime}\sum_{m\neq
0}\frac{1}{m}\alpha_{m}^{\mu}\alpha_{-m}^{\nu},$ (E.6)

and then the commutators

$\displaystyle\frac{1}{{\cal N}}[\hat{M}^{\mu\nu},x^{\lambda}]=$
$\displaystyle\alpha^{\prime}\delta^{[\mu}x_{0}^{\nu]},$
$\displaystyle\frac{1}{{\cal N}}[\hat{M}^{\mu\nu},\alpha_{l}^{\lambda}]=$
$\displaystyle 2\alpha^{\prime}\delta^{[\mu}\alpha_{l}^{\nu]},$ (E.7)

which show explicitly what stated since the two variations differ by a factor
$2$.

We can evaluate the same using the fields. To this aim one has to go back to
the original definitions and compute

$\displaystyle\frac{1}{{\cal N}}[\hat{M}^{\mu\nu}$
$\displaystyle,L^{\lambda}(w)]=$ $\displaystyle=$
$\displaystyle\oint_{z=0,|z|>|w|}\frac{dz}{2\pi
i}\left(L^{\mu}_{(\neq)}(z)\partial L^{\nu}(z)-L^{\nu}_{(\neq)}(z)\partial
L^{\mu}(z)\right)\,L^{\lambda}(w)$
$\displaystyle-L^{\lambda}(w)\,\oint_{z=0,|z|<|w|}\frac{dz}{2\pi
i}\left(L^{\mu}_{(\neq)}(z)\partial L^{\nu}(z)-L^{\nu}_{(\neq)}(z)\partial
L^{\mu}(z)\right),$ (E.8)

and carefully use eq.s (D.5).

The proper Lorentz generator is then

$M^{\mu\nu}=\hat{M}^{\mu\nu}+\frac{1}{2}(x^{\mu}_{0}p^{\nu}_{0}-x_{0}^{\nu}p_{0}^{\mu}).$
(E.9)

###  E.2 Generators $\underline{M}^{ij}$

It is immediate to compute

$\displaystyle[\underline{M}^{ij},\,{\underline{L}}^{k}]$
$\displaystyle=2i\eta^{k[i}{\underline{L}}^{j]},$
$\displaystyle[\underline{M}^{ij},\,{\underline{L}}^{+}]$ $\displaystyle=0,$
$\displaystyle[\underline{M}^{ij},\,{\underline{L}}^{-}]$ $\displaystyle=0,$
(E.10)

so that

$\displaystyle[\underline{M}^{ij},\,{\underline{A}}^{k}_{n}]$
$\displaystyle=2i\eta^{k[i}{\underline{A}}_{n}^{j]},$
$\displaystyle[\underline{M}^{ij},\,{\underline{A}}_{n}^{-}]$
$\displaystyle=[\underline{M}^{ij},\,\tilde{\underline{A}}_{n}^{-}]=0.$ (E.11)

Then we are led to the expression for the physical states

$\underline{M}^{ij}|_{DDF}=i\sum_{n\neq
0}\frac{1}{n}{\underline{A}}_{n}^{i}{\underline{A}}^{j}_{-n}.$ (E.12)

This expression reproduces the desired commutation relations for
${\underline{A}}_{n}$ with $n\neq 0$ but fails for ${\underline{A}}_{0}$. Let
us see whether it is not possible to add some zero modes contributions to fix
this issue. For example the natural guess
$2{\underline{x}}_{0}^{[i}{\underline{p}}_{0}^{j]}$ fails. In facts

$[2{\underline{x}}_{0}^{[i}{\underline{p}}_{0}^{j]},\,{\underline{A}}^{k}_{m}]=2i\eta^{k[i}{\underline{A}}_{0}^{j]},$
(E.13)

for all $n$ because of the ${\underline{x}}_{0}$ zero mode interacts with
$\partial{\underline{L}}$. But this failure can be compensated as

$\displaystyle[-2i\sum_{n\neq
0}\frac{1}{n}{\underline{A}}_{n}^{[i}{\underline{A}}_{0}^{j]},\,{\underline{A}}^{k}_{m}]=$
$\displaystyle-2i\eta^{k[i}{\underline{A}}_{0}^{j]},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ m\neq 0.$ (E.14)

The final expression which is valid also for ${\underline{A}}_{0}^{i}$ is then

$\displaystyle\underline{M}^{ij}=$ $\displaystyle i\sum_{n\neq
0}\frac{1}{n}{\underline{A}}_{n}^{i}{\underline{A}}^{j}_{-n}$
$\displaystyle-2i\sum_{n\neq
0}\frac{1}{n}{\underline{A}}_{n}^{[i}{\underline{A}}_{0}^{j]}+\frac{2}{\sqrt{2\alpha^{\prime}}}{\underline{x}}_{0}^{[i}{\underline{A}}_{0}^{j]}.$
(E.15)

###  E.3 Generators $\underline{M}^{+i}$

It again is immediate to compute

$\displaystyle[\underline{M}^{+i},\,{\underline{L}}^{k}]$
$\displaystyle=-i\delta^{ki}{\underline{L}}^{+},$
$\displaystyle[\underline{M}^{+i},\,{\underline{L}}^{+}]$ $\displaystyle=0,$
$\displaystyle[\underline{M}^{+i},\,{\underline{L}}^{-}]$
$\displaystyle=-i{\underline{L}}^{i},$ (E.16)

so that

$\displaystyle[\underline{M}^{+i},\,{\underline{A}}^{k}_{n}]$
$\displaystyle=-i\delta^{ki}{\underline{A}}_{0}^{+}\,\delta_{n,0},$
$\displaystyle[\underline{M}^{+i},\,{\underline{A}}_{n}^{-}]$
$\displaystyle=-i{\underline{A}}^{i}_{n},$
$\displaystyle[\underline{M}^{+i},\,\tilde{\underline{A}}_{n}^{-}]$
$\displaystyle=0.$ (E.17)

Then we are led to the expression for the physical states which do not require
the use of DDF zero modes ${\underline{A}}_{0}$

$\underline{M}^{+i}|_{DDF}=0.$ (E.18)

Notice that this expression is valid only when using ${\underline{A}}^{i}$ and
the improved Brower operators $\tilde{\underline{A}}^{-}$. This means that
when we have an expression with ${\underline{A}}^{-}$ it has to be reexpressed
using $\tilde{\underline{A}}^{-}$ and ${\cal L}$: this observation is
important for the computation of the expectation values of the Casimir
operators of the different physical states obtained by the DDF operators.

###  E.4 Generator $\underline{M}^{+-}$

It again is immediate to compute

$\displaystyle[\underline{M}^{+-},\,{\underline{L}}^{k}]$ $\displaystyle=0,$
$\displaystyle[\underline{M}^{+-},\,{\underline{L}}^{+}]$
$\displaystyle=+i{\underline{L}}^{-},$
$\displaystyle[\underline{M}^{+-},\,{\underline{L}}^{-}]$
$\displaystyle=-i{\underline{L}}^{+},$ (E.19)

so that

$\displaystyle[\underline{M}^{+-},\,{\underline{A}}^{k}_{n}]$
$\displaystyle=0,$
$\displaystyle[\underline{M}^{+-},\,{\underline{A}}_{n}^{-}]$
$\displaystyle=-i{\underline{A}}^{-}_{n},$
$\displaystyle[\underline{M}^{+-},\,\tilde{\underline{A}}_{n}^{-}]$
$\displaystyle=-i\tilde{\underline{A}}_{n}^{-}.$ (E.20)

Notice that these results rely on the invariance of
$\exp\left(i\frac{n}{\alpha^{\prime}}\frac{{\underline{L}}^{+}}{{\underline{p}}_{0}^{+}}\right)$.

Exactly as for the generators $\underline{M}^{+i}$ it cannot be represented
using ${\underline{A}}^{i}$ and $\tilde{\underline{A}}^{-}$, not even in the
physical sector. The reason is that it can be only a functional of
$\tilde{\underline{A}}_{n}^{-}$ since $\underline{M}^{+-}$ commutes with all
${\underline{A}}^{i}$ but $\tilde{\underline{A}}_{n}^{-}$ satisfy a Virasoro
algebra and therefore

###  E.5 Generators $\underline{M}^{-i}$

These are the most interesting generators since they combine a true rotation
and a compensating conformal transformation in the transverse directions.

It again is immediate to compute

$\displaystyle[\underline{M}^{-i},\,{\underline{L}}^{k}]$
$\displaystyle=-i\delta^{ki}{\underline{L}}^{-},$
$\displaystyle[\underline{M}^{-i},\,{\underline{L}}^{+}]$
$\displaystyle=-i{\underline{L}}^{i},$
$\displaystyle[\underline{M}^{-i},\,{\underline{L}}^{-}]$ $\displaystyle=0.$
(E.21)

Because of the non trivial rotation of ${\underline{L}}^{+}$ is not immediate
to compute the effect on DDF and Brower operators.

The final result is that

$\displaystyle[\underline{M}^{-i}(E),\,{\underline{A}}^{k}_{n}(E)]=$
$\displaystyle i\frac{n}{{\underline{\alpha}}^{+}_{0}}\sum_{l\neq
n}\frac{1}{n-l}{\underline{A}}^{k}_{l}(E)\,{\underline{A}}^{i}_{n-l}(E)+\frac{n}{({\underline{\alpha}}^{+}_{0})^{2}}\underline{M}^{+i}(E){\underline{A}}^{k}_{n}(E)$
$\displaystyle+\delta^{ki}\left[-\frac{i}{2{\underline{\alpha}}_{0}^{+}}\sum_{l}{\underline{A}}^{j}_{l}(E)\,{\underline{A}}^{j}_{n-l}(E)-i{\tilde{\underline{A}}}^{-}_{n}(E)\right].$
$\displaystyle=$ $\displaystyle
i\frac{n}{{\underline{\alpha}}^{+}_{0}}\sum_{l\neq
n}\frac{1}{n-l}{\underline{A}}^{k}_{l}(E)\,{\underline{A}}^{i}_{n-l}(E)+\frac{n}{({\underline{\alpha}}^{+}_{0})^{2}}\underline{M}^{+i}(E){\underline{A}}^{k}_{n}(E)-i\delta^{ki}{{\underline{A}}}^{-}_{n}(E),.$
(E.22)

The expression for the variation of ${\underline{A}}^{k}$ under
$\underline{M}^{-i}(E)$ may seem quite strange since it involves
$\underline{M}^{+i}(E)$ but a simple consistency check is immediate to
perform: the scaling w.r.t. $\underline{M}^{+-}(E)$ works because of
$1/({\underline{\alpha}}_{0}^{+})^{2}$. The same is true for the first term.

The previous result is obtained first computing the variation of
${\underline{A}}^{k}$ then inferring the expression of the generator
$\underline{M}^{-i}$ and then using this expression we can compute the
variation of ${\underline{A}}^{-}$ and $\tilde{\underline{A}}^{-}$. The
expression of the generator $\underline{M}^{-i}$ restricted to DDF operators
is

$\displaystyle\underline{M}^{-i}|_{DDF}=$ $\displaystyle i\sum_{m\neq
0}\frac{1}{m}{\underline{A}}_{m}^{-}(E){\underline{A}}^{i}_{-m}(E)-\frac{1}{({\underline{\alpha}}_{0}^{+})^{2}}\underline{M}^{+i}(E){\cal
L}_{0}(E)$ $\displaystyle=$ $\displaystyle i\sum_{m\neq
0}\frac{1}{m}{\tilde{\underline{A}}}_{m}^{-}(E){\underline{A}}^{i}_{-m}(E)+\frac{i}{2{\underline{\alpha}}_{0}^{+}}\sum_{m\neq
0}\frac{1}{m}{\cal
L}_{m}(E){\underline{A}}^{i}_{-m}(E)-\frac{1}{({\underline{\alpha}}_{0}^{+})^{2}}\underline{M}^{+i}{\cal
L}_{0}(E),$ (E.23)

where ${\cal L}_{n}$ is defined in (2.13). The first term in the r.h.s. of the
first line gives to the first and third terms of the final form of the
${\underline{A}}^{k}(E)$ variation.

Notice that there are apparently normal ordering problems since
${\underline{A}}_{m}^{-}(E)$ and ${\underline{A}}^{i}_{-m}(E)$ do not commute,
however the ambiguity is proportional to the zero modes which are not DDF
operators.

##  Appendix F Details on extracting Poincaré irreps from Lorentz irreps

In this appendix we would like to give some details on how to perform the
splitting of Lorentz polarizations into Poincaré irreps. This is useful if we
want to compute the Casimirs in the covariant formalism.

We consider massive states with mass $M$ and perform the decomposition in rest
frame and then we covariantize the result.

We use $I,J,\dots=1,2,\dots D-1$.

###  F.1 Lorentz vector splits

Let us consider the Lorentz vector $T_{\mu}$. In the rest frame from the
$SO(D-1)$ point of view we have actually two irreps

$T_{\mu}\rightarrow T_{I}\,\oplus\,T_{0},$ (F.1)

which have two different Casimirs.

This split seems to break Lorentz invariance but it is nor so. We have in fact
the following covariant characterization of the two irreps555 Remember that in
rest frame $k^{0}=-k_{0}=M$.

$\displaystyle T_{0}:$ $\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\
t=\frac{k^{\mu}}{M}T_{\mu}\neq 0,\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ $ $\displaystyle T$
${}_{\mu}^{(\cdot)}=\frac{k^{\mu}}{M}t\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ $
$\displaystyle\rightarrow$ $\displaystyle\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \mbox{not
transverse vector and proportional to }k_{\mu}$ $\displaystyle T_{I}:$
$\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\
k^{\mu}t_{(\ydiagram{1})\mu}=0,\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ $ $\displaystyle t$
${}_{(\ydiagram{1})\mu}=T_{\mu}+\frac{k^{\mu}}{M}t$ $\displaystyle\rightarrow$
$\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{transverse vector}.$ (F.2)

Finally we can write the decomposition of the Lorentz vector into Poincaré
irreps as

$T_{\mu}=t_{(\ydiagram{1})\mu}+\frac{k^{\mu}}{M}t.$ (F.3)

###  F.2 Lorentz antisymmetric two tensor splits

Let us consider the Lorentz antisymmetric tensor $A_{[\mu\nu]}$. From the
$SO(D-1)$ point of view in the rest frame we have actually two irreps

$A_{[\mu\nu]}\rightarrow A_{[IJ]}\,\oplus\,A_{[I0]},$ (F.4)

which again have two different Casimirs.

The covariant characterization is

$\displaystyle A_{[I0]}:$ $\displaystyle\leavevmode\nobreak\
\leavevmode\nobreak\
a_{(\ydiagram{1})\mu}=\frac{k^{\mu}}{M}A_{\mu\nu},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\
k^{\mu}a_{(\ydiagram{1})\mu}=0,\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ $ $\displaystyle A$
${}_{(\ydiagram{1})\mu\nu}=-2\frac{1}{M}k_{[\mu}a_{(\ydiagram{1})\nu]}$
$\displaystyle A_{IJ}:$ $\displaystyle\leavevmode\nobreak\
\leavevmode\nobreak\ k^{\mu}a_{(\ydiagram{1,1})\mu\nu}=0,\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ $ $\displaystyle a$
${}_{(\ydiagram{1,1})\mu\nu}=A_{\mu\nu}-A_{(\ydiagram{1})\mu\nu}.$ (F.5)

All these relations can be verified either directly or in rest frame.

Finally we can write the decomposition of the Lorentz antisymmetric two tensor
into Poincaré irreps as

$A_{\mu\nu}=a_{(\ydiagram{1,1})\mu\nu}-2\frac{1}{M}k_{[\mu}a_{(\ydiagram{1})\nu]}.$
(F.6)

####  F.2.1 Lorentz symmetric traceless two tensor splits

Let us consider the Lorentz symmetric traceless tensor $S_{\mu\nu}$. The
traceless condition means that

$S_{00}=S_{KK}.$ (F.7)

From the $SO(D-1)$ point of view in the rest frame we have actually three
irreps

$S_{\mu\nu}\rightarrow\left(S_{IJ}-\frac{1}{D-1}\delta_{IJ}S_{KK}\right)\,\oplus\,S_{I0}\,\oplus\,S_{00}\equiv
S_{KK},$ (F.8)

which have once again three different Casimirs.

The covariant characterization of the lowest irreps is easy and reads

$\displaystyle S_{00}:$ $\displaystyle\leavevmode\nobreak\
\leavevmode\nobreak\
s_{(\cdot)}=\frac{k^{\mu}}{M}\frac{k^{\nu}}{M}S_{\mu\nu},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ $ $\displaystyle S$
${}_{(\cdot)\mu\nu}=\frac{k^{\mu}}{M}\frac{k^{\nu}}{M}s_{(\cdot)}$
$\displaystyle S_{0I}:$ $\displaystyle\leavevmode\nobreak\
\leavevmode\nobreak\
s_{(\ydiagram{1})\mu}=\frac{k^{\nu}}{M}S_{\mu\nu}+\frac{k_{\mu}}{M}s_{(\cdot)},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\
k^{\mu}s_{(\ydiagram{1})\mu}=0,\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ $ $\displaystyle S$
${}_{(\ydiagram{1})\mu\nu}=\frac{k_{(\mu}s_{(\ydiagram{1})\nu)}}{M}.$ (F.9)

Notice that not all tensors are traceless, i.e.

$S_{(\cdot)\mu\mu}\neq 0,\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ S_{(\ydiagram{1})\mu\mu}=0.$ (F.10)

While the highest irrep is a little trickier since one would wrongly guess
that the last irrep is the left over

$S_{IJ}:\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
s_{(\ydiagram{2})\mu\nu}=S_{\mu\nu}-S_{(\ydiagram{1})\mu\nu}-S_{(\cdot)\mu\nu},$
(F.11)

but this is wrong since it misses $\frac{1}{D-1}\delta_{IJ}S_{KK}$ in the
irrep definition. The proper definition is

$S_{IJ}-\frac{1}{D-1}\delta_{IJ}S_{00}:\leavevmode\nobreak\
\leavevmode\nobreak\
s_{(\ydiagram{2})\mu\nu}=S_{\mu\nu}-S_{(\ydiagram{1})\mu\nu}-S_{(\cdot)\mu\nu}-\frac{1}{D-1}P_{\mu\nu}s_{(\cdot)},$
(F.12)

with $P_{\mu\nu}=\eta_{\mu\nu}-\frac{k_{\mu}k_{\nu}}{k^{2}}$ so that
$P_{\mu\nu}k^{\nu}=0$ and $P^{\lambda}_{\lambda}=D-1$. This expression can be
verified in the rest frame and making use of $S_{00}=S_{KK}$. It has the
wanted properties

$\displaystyle k^{\mu}s_{(\ydiagram{2})\mu\nu}=0,\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ s_{(\ydiagram{2})\rho\rho}=0.$
(F.13)

Finally we can write the decomposition of the Lorentz symmetric traceless two
tensor into Poincaré irreps as

$S_{\mu\nu}=s_{(\ydiagram{2})\mu\nu}-2\frac{1}{M}k_{(\mu}s_{(\ydiagram{1})\nu)}+\left(\frac{1}{D-1}P_{\mu\nu}+\frac{k_{\mu}k_{\nu}}{M^{2}}\right)s_{(\cdot)}.$
(F.14)

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# KeyDetect - Detection of anomalies and user based on Keystroke Dynamics

Soumyatattwa Kar
<EMAIL_ADDRESS>Department of Mechanical Engineering, Carnegie Mellon
University, 5000 Forbes Ave, Pittsburgh, PA 15213 Abhishek
Bamotra11footnotemark: 1
<EMAIL_ADDRESS>Correspondence author 412-692-1807 Bhavya Duvvuri
<EMAIL_ADDRESS>Department of Civil and Environmental Engineering,
Carnegie Mellon University, 5000 Forbes Ave, Pittsburgh, PA 15213 Radhika
Mohanan11footnotemark: 1
<EMAIL_ADDRESS>
###### Abstract

Cyber attacks has always been of a great concern. Websites and services with
poor security layers are the most vulnerable to such cyber attacks. The
attackers can easily access sensitive data like credit card details and social
security number from such vulnerable services. Currently to stop cyber
attacks, various different methods are opted from using two-step verification
methods like One-Time Password and push notification services to using high-
end bio-metric devices like finger print reader and iris scanner are used as
security layers. These current security measures carry a lot of cons and the
worst is that user always need to carry the authentication device on them to
access their data. To overcome this, we are proposing a technique of using
keystroke dynamics (typing pattern) of a user to authenticate the genuine
user. In the method, we are taking a data set of 51 users typing a password in
8 sessions done on alternate days to record mood fluctuations of the user.
Developed and implemented anomaly-detection algorithm based on distance
metrics and machine learning algorithms like Artificial Neural networks (ANN)
and convolutional neural network (CNN) to classify the users. In ANN, we
implemented multi-class classification using 1-D convolution as the data was
correlated and multi-class classification with negative class which was used
to classify anomaly based on all users put together. We were able to achieve
an accuracy of 95.05% using ANN with Negative Class. From the results
achieved, we can say that the model works perfectly and can be bought into the
market as a security layer and a good alternative to two-step verification
using external devices. This technique will enable users to have two-step
security layer without worrying about carry an authentication device.

##  1 Introduction

Cyber security has always been the most important concern about the internet
usage. Everyday millions of people get attacked by impostors to extract their
valuable information like credit card details, social security number etc.
These information are vital to a person and hold a lot of value, exploitation
of such information can result in great loss [13]. Also, cyber security is a
must and companies spend millions of dollar on improvement and constant
development of security layers to protect sensitive data from getting
attacked.

There are various type of security layers deployed in the internet sphere some
of them are two-step verification using One-Time Password (OTP), push
verification etc[10]. In such security layers, user is required to enter the
credentials for the account, select two-step verification and wait for a
message which will contain OTP or a push notification on their mobile. Using
the message/notification, the user verifies the login. Such security layers
are good in protecting the data even when an imposter is able to extract
user’s credentials. These layers are strong and helpful but requires a user to
always carry their authentication device on them. So, loss or unavailability
of the authentication device will result in blocking the user from accessing
the data.

To tackle this issue, high-end security devices like finger print reader and
iris scanners are used[2]. One proposed approach is using keystroke dynamics
of a person to authenticate the user as second layer security[6][5]. The
technique uses the user’s unique typing pattern and distinguish the genuine
user from an imposter. Using keystroke dynamics, even if the imposter knows
the credentials of a person, imposter won’t be able to login as keystroke
pattern of an imposter will vary significantly from the genuine user[1][7].
The dataset used in this problem is taken on a physical computer keyboard and
can also be implemented on smartphones[15].

We have tested various anomaly-detection algorithms based on distance metrics
and machine learning algorithms like Random forest, Support vector machine
(SVM), Artificial Neural networks (ANN) and convolutional neural network (CNN)
to detect genuine users (discussed in a later section). Accuracy or
performance of an algorithm is the measure one uses to evaluate an algorithm,
if the results from an algorithm makes sense or are just random outputs with
minimal or no sense. To evaluate our distance metrics algorithms, we are using
European standard norm of false-alarm rate and miss rate[8]. Whereas for
machine learning algorithms, we are giving accuracy of the classifiers to
evaluate the algorithms.

The results demonstrated by the CNN 1-D convolution and ANN with negative
class algorithms (94.6% and 95.05% respectively) reflect how effectively the
keystroke dynamics can be used in authenticating a genuine user.

##  2 Related work

Keystroke dynamics has been taken in consideration from a long time and a lot
of research has been done on the topic. In 2009, Kevin Killourhy and Roy
Maxion have collected the benchmark data set for keystroke dynamics using
windows PC and physical keyboard to record the timing vector of a particular
password for 51 users. Many researchers have previous worked on this data set
to extract the most information. In 2009, Kevin and Roy themselves tried to
perform anomaly-detection algorithms to distinguish the imposters from the
genuine users. The results demonstrated by the algorithms were not up to the
expectations, the ANN algorithm worked the worst.

In late 2009, a research was conducted on dynamic keystroke in mobile phone to
identity users[15]. They used fuzzy classifier which resulted in
distinguishing 3 different features (key hold time, time difference between
pressing and releasing a key and time difference between release and pressing
a key). They tried different algorithms like Naive Bayes, Back Propagation
Neural Network, K start, Radial Basis Function Network.

Later, a research was done on user’s emotional state classification[3]. The
data set for this research was collected by the research group. The aim of the
work was to classify a user’s mood into emotional classes like confidence,
relaxation, sadness, anger, excitement tiredness and hesitance. Different
machine learning classifiers were used and resulted in a max accuracy of 88%.

There has been research in age-group classification[11] and checking for
fatigue based [4] on typing pattern of a user. There are various other
research going on based on keystroke dynamics in security and other user
specific classification. The most interesting research ouput is by Yohan
Muliono, Hanry Ham and Dion Darmawan. The aim of the work was to classify
users based on keystroke dynamics of the users. The group performed different
machine learning algorithms like Support vector machine with linear, RBF and
Poly kernel, standard deep learning and deep learning algorithm with modified
adam optimizer. The group was able to reach a max accuracy of 92.6% which was
the mark to improve.

##  3 Data

###  3.1 Data collection details

The data-set used for the various models was taken from the paper [6]. It
consists of typing samples of 51 users each typing 400 repetitions of the
password ”.tie5Roanl”. The password selected was representative of typical,
strong password. The data was collected in 8 data-collection sessions (50
passwords for each), with at least one day between the sessions, to capture
some of the day-to-day variation of each subject’s typing. The data is a point
in p-dimensional space, where p is the number of features in the timing
vectors. So, training data is a cloud of points.

###  3.2 Subject details

The set of subjects/users consisted of 30 males and 21 females, 8 left-handed
and 43 right-handed.The median age group was 31-40, the youngest was 18-20 and
oldest was 61-70. The subjects’ took between 1.25 and 11 minutes, with the
median session taking about 3 minutes.

###  3.3 Data-set Features

The data consists of timing features of users represented as keydown-keydown
times, keyup-keydown times, hold times and Enter key times. Total number of
timing features extracted were 31, they were stored in seconds(as floating-
point numbers).

Figure 1: Timing features considered in the data set

###  3.4 Data pre-processing

Data was inspected for outliers, missing values and noise values. Ouliers were
found in the data set which were removed. The data did not have any missing
values or noise and it was found to be normally distributed.

Figure 2: Data Distribution Figure 3: Outlier Detection

##  4 Methods

In our data-set we had total of 51 users. Our goal was to implement an anomaly
detection system and then extend the same to multi-class classification
problem so that we are able to identify user only by using keystroke data.

By using neural networks we have mainly concentrated our efforts in three
categories. First, develop a model for multi-class classification problem
where we have tried to classify one user among the 51 given use. Second we
have tried to detect an anomalous user which the model has never seen before
during training using neural network. Third we tried to optimize our models
for better accuracy and performance.

For multi class classification our approach is to first try fully connected
neural network with two hidden layers. Next tune the model by varying the no
of nodes, learning rates and epoch to find optimum hyper parameters. Second we
try 1-dimensional convolution layers. Since during typing of password one key
is pressed after the other it made logical sense to try to capture the spatial
relation, if there were any, in the features (which are keystroke timings) by
using 1D convolution layers. Next tune the model by varying number of layers,
kernel size and channels in each layers learning rate to find optimum hyper
parameters.

In the next section we will try to implement the concept of negative class for
anomaly detection. This will just be an experimentation to see if we are able
detect an user the model has never seen before and categorize it as an
anomalous user. In this approach instead of using all the 51 users for
training a classification model we will use only 31 users. The 31 users
represent 31 classes. One more class, called the negative class, is created
which has random samples from the remaining (51-31) 20 users. Our approach is
to see if the model correctly classifies an anomalous user to the negative
class.

Finally we improve the model performance by changing the hyper parameters like
no of nodes, no of channels, kernel size. We will also be implementing a
learning rate scheduler to schedule a learning rate decay when the validation
loss plateau.

##  5 Experiments

We implemented all the best performing algorithms from the above cited paper
and going further using our learning from NN based object removal[12]
implemented neural network, SVM and random forest. We describe the models that
we implemented in in this section.

###  5.1 Euclidean

It is a anamoly detection model. Training data is used to get the centre of
the cloud, which is a mean vector. The anomaly score of the test vector is
based on its proximity to this mean vector, calculated as the squared
Euclidean distance between the test vector and the mean vector.

###  5.2 Manhattan

The model works similar to the Euclidean model, expect that the distance
metric used is manhattan, where the distance between two points is measured
along axes at right angles.

###  5.3 Manhattan (Scaled)

As describe in the paper that we referred, anomaly score is calculated as

$\sum_{i=1}^{p}2^{|x_{i}-y_{i}|/a_{i}}=1$

where xi and yi are the i-th features of the test and mean vectors
respectively, and ai is the average absolute deviation from the training
phase.The score resembles a Manhattan-distance calculation, except each
dimension is scaled by ai.

###  5.4 Mahalanobis

In this model, during the training phase, mean vector and co-variance matrix
of variables are calculated. The mahalanobis is calculated as

$(x-y)^{T}*S^{-1}*(x-y)$

where S is the covariance matrix

###  5.5 Mahalanobis (Normed)

This model works the same way as Mahalanobis, except that the anomaly score is
calculated by “normalizing” the Mahalanobis distance by dividing the distance
by

$||x||*||y||$

###  5.6 Z-score

In this model, the detector calculates the mean and standard deviation of each
timing feature in training phase. In the test phase, the detector computes the
absolute z-score of each feature of the test vector. The z-score for the i-th
feature is calculated as

$|x_{i}-y_{i}|/s_{i}$

where $x_{i}$ and $y_{i}$ are the i-th features of the test and mean vectors
respectively and $s_{i}$ is the standard deviation from the training phase.
The anomaly score is a count of how many z-scores exceed a threshold.

###  5.7 SVM (one-class)

In this algorithm, the training data is projected into high dimensional space
and linear separator is defined. The anomaly score is calculated as distance
between linear separator and test vector that is projected in the same high
dimensional space. The hyperparameter v was optimised.

###  5.8 Support Vector Machine (SVM)

It is a multiclass classification problem, that we are implementing for user
identifictaion. The intuition behind the multiclass SVM is that, if our
classification rule is

$y_{i}=\textrm{argmax}_{j}\;h(x_{i})*w_{j}+w_{0,j{{}^{\prime}}}$

we should simply make sure that if

$y_{i}=j,\textrm{then }h(x_{i})*w_{j{{}^{\prime}}}+w_{0,j{{}^{\prime}}}$

is greater than

$h(x_{i})*w_{j{{}^{\prime}}}+w_{0,j{{}^{\prime}}}$

for all

$j{{}^{\prime}}=j$

by the largest margin, in the same way that we make sure that

$h(x_{i})*w_{j}+w_{0,j}\geq 1$

in the binary SVM.So we can directly optimize over all of our decision
boundaries with constraints that enforce it, and the same objective as before
(but now summed over all decision boundaries):

$min_{w1,...,wL1,w0,1,...,w0,Ly,s1,...,sN}\frac{1}{2}\sum_{j=1}^{L_{y}}{||w_{j}||}+\lambda\sum_{j=1}^{N}{s_{i}}$

Figure 4: ROC curve of subject 38 using mahalanobis normed

The terminologies of ROC Curve are explained below:

  * •

Miss rates - Percentage of impostor passwords that are not detected

  * •

Hit rate - Frequency with which impostors are detected (i.e., 1 - miss rate)

  * •

False-alarm rate - Frequency with which genuine users are mistakenly detected
as impostors.

Figure 5: Nerual Network architecture with 1-D convolutional layers

###  5.9 Neural Network

For the calculation of equal-error rate, we chose a threshold so that the
detector’s miss and false-alarm rates are equal. Zero-miss false-alarm rate is
calculated by choosing the threshold so that the false-alarm rate is minimized
under the constraint that the miss rate be zero.

Experimentation of Neural networks were carried out in two phases. First
implementation of fully connected network and secondly implementation of 1
dimensional convolutional network.

In fully connected network implementation we have implemented 1 hidden layer
and 2 hidden layers configuration for the multi-class classification problem .
We also varied the number of nodes in each of the layers to find the most
optimum configuration. Based on our test results the best configuration was 2
hidden layers with 80 nodes in first layer and 60 nodes in second layer. The
accuracy obtained from the model was 92.2 % . Learning rate scheduler was used
to decrease learning rate with a decay factor of 0.1 when validation loss
plateaued.

In 1D convolution we implemented multiple combinations of convolutional layers
followed by fully connected layers. After the carrying out hyper parameter
tuning the best model configuration was found to be 2 1D convolutional layers
back to back. The first layer has 16 channels with kernel size 3. The second
convolutional layer has 32 channels with kernel size 3. This is followed by 2
fully connected layers having 992 nodes and 128 nodes. The final layer has 51
nodes corresponding to 51 classes. While training the model a learning rate
scheduler was used to decrease the learning rate by a factor of 0.1 when
validation loss plateaued. We were able to achieve an accuracy of 94.6%. This
is a more than 2% accuracy increase than the fully connected configuration.
This clearly shows that there is indeed some spatial correlation among the
features which has been captured in the 1D convolutional layers. during trying
out various configurations of the network we have also noticed that as we
increased the model complexity by increasing the number of convolutional
layers or the number of channels in each layer the accuracy drops.

Figure 6: Confusion matrix for ANN with Negative Class

Using the above mentioned model configuration we implemented the negative
class approach in an attempt to do anomaly detection. In this approach we have
taken 31 users of the total 51 users as 31 classes. Next we created one more
class, called the Negative Class that is a random combination of all remaining
(51-31) 20 classes. Next we train the model using a total of (31+1) 32
classes. Our goal is to see if the trained model is able to classify an
anomalous user (which is a data point that doesn’t belong to any of the 31
users used to train model) successfully in the negative class. After training
the model we were able to get an overall model accuracy of 95.05% and
confusion matrix is shown in figure 6. If we consider the negative class with
respect to all other classes, the recall is 80.9%, precision is 67.22%,
f-score is 0.733.

###  5.10 Random Forest Classifier

For classifying the users into different classes we used Random Forest
Classifier. It is an ensemble method of decision trees generated on a randomly
split dataset. The dataset was split in the ratio of 75:25 where 75% was used
for training the model and remaining 25% for testing.Decision trees were
constructed for each sample, of which prediction results were obtained.The
results with maximum vote was considered as the final prediction. The accuracy
of this model on the test data is around 93.66%

##  6 Performance

The performance matrices of the anomaly-detection algorithms based on distance
metrics has been split into Equal Error Rate and Zero-Miss False-Alarm Rate.
The results achieved by the algorithms has been listed in the descending order
of the average of the parameter. Average Error Rate and standard deviation of
Error Rate is shown in Table 1. Average and Deviation of Zero-Miss False-Alarm
Rate is shown in Table 2. From the results, we can say that the distance
metrics algorithms are not performing well on anomaly detection.

Table 1: Table of Average and Standard Deviation (S.D.) of Equal Error Rate (EER). Detector | Average EER | S.D. EER
---|---|---
Euclidean | 0.265 | 0.079
Manhattan | 0.206 | 0.077
Mahalanobis | 0.193 | 0.101
Mahalanobis (Normed) | 0.166 | 0.092
Manhattan (Scaled) | 0.141 | 0.068
Z-Score | 0.135 | 0.065
Table 2: Table of Average and Standard Deviation (S.D.) of Zero-Miss False-Alarm Rate (ZFR). Detector | Average ZFR | S.D. ZFR
---|---|---
Z-Score | 0.535 | 0.275
Manhattan (Scaled) | 0.540 | 0.271
Mahalanobis | 0.645 | 0.299
Mahalanobis (Normed) | 0.666 | 0.301
Manhattan | 0.711 | 0.224
Euclidean | 0.753 | 0.212

In table 3, we can see the performance of our machine learning classifier
algorithms. From the table, we can say Artificial Neural Network with negative
class classification performs the best with an accuracy of 95.05%.

Table 3: Table of Accuracy of Machine Learning Models. Multi-Class Classification | Accuracy %
---|---
ANN (With Negative Class) | 95.05
ANN (1-D Conv.) | 94.6
Random Forest | 93.66
ANN (Fully Connected) | 92.2
Support Vector Machine | 75.6

##  7 Conclusion

The results presented here clearly shows that based on the feature vector
created from keystroke dynamics it is possible to classify users. Also,
results obtained from training neural networks of different configurations we
can see that there are some spatial correlation in the feature vector. However
the data collected was based on one password that was typed by 51 users. An
extension of this work could be to analyse the keystroke dynamics of multiple
users over some commonly occurring words in the English language tokened as
Flag Words. Once a user starts using the keyboard the keystroke dynamics can
be monitored online and in the event any flag word in typed in, the data can
be used to verify the authenticity of the user. This feature can be used in
smartphones, desktop, laptops or any other systems having a keyboard a an
extra layer of security of user identification. As a future work, emotion
detection based on keystrokes tactile sensing can be added as extra layer of
security, if highly sensitive soft-bodied tactile[14, 9] sensors are used.

## References

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